Hmm, some minor mistakes from being 11 hours driving through California. Apparently, the speed in the original coordinate system was $$u$$ not $$\beta c$$ and this allows $$v$$ to be used as the Lorentz parameter after all. Also I found a typo.
---
Also in proper language, we can describe these events and trajectories in coordinates. In the inertial coordinates (x,t) where d and e are the standard of rest:
$$x_a - x_O = c ( t_a - t_O ) \\ x_b - x_O = u ( t_b - t_O ) \\ x_d - x_O = 0 \\ x_e - x_O = L $$
where L is the distance between d and e as measured in this frame, which makes much more sense for a name than "x".
To find the coordinates of $$A = a \cap e$$, we solve:
$$x_a - x_O = c ( t_a - t_O ) \\ x_e - x_O = L \\ x_A = x_a = x_e \\ t_A = t_a = t_e$$
with solution $$x_A = x_O + L , \; t_A = t_O + \frac{L}{c}$$
Likewise, the coordinates of $$B = b \cap e$$ are $$x_B = x_O + L , \; t_B = t_O + \frac{L}{u}$$
Now to check if the Lorentz equations are self-consistent, we need to transform a, b, d,and e into new coordinates and see if they give the same solution as directly transforming A and B. (We're using the language of algebra, as did Einstein in 1905, but in 1908 Minkowski demonstrated that one can also use the language of geometry which provides deeper math reasons for the basic self-consistency of Special Relativity.)
$$t = \frac{t' + \frac{v}{c^2} x'}{\sqrt{1 - \frac{v^2}{c^2}}} \\ x = \frac{x' + v t'}{\sqrt{1 - \frac{v^2}{c^2}}}$$
Step 1: Find the (original system) coordinates for O, A and B. I will illustrate the general solution with O:
$$t_O = \frac{t'_O + \frac{v}{c^2} x'_O}{\sqrt{1 - \frac{v^2}{c^2}}} \\ x_O = \frac{x'_O + v t'_O}{\sqrt{1 - \frac{v^2}{c^2}}}$$
Then
$$t_O - \frac{v}{c^2} x_O = \frac{t'_O + \frac{v}{c^2} x'_O - \frac{v}{c^2} x'_O - \frac{v}{c^2} v t'_O}{\sqrt{1 - \frac{v^2}{c^2}}} = t'_O \frac{1 - \frac{v^2}{c^2}}{\sqrt{1 - \frac{v^2}{c^2}}} = t'_O \sqrt{1 - \frac{v^2}{c^2}}$$
Thus $$t'_O = \frac{t_O - \frac{v}{c^2} x_O}{\sqrt{1 - \frac{v^2}{c^2}}}$$
Similarly: $$x'_O = \frac{x_O - v t_O}{\sqrt{1 - \frac{v^2}{c^2}}}$$
Likewise,
$$t'_A = \frac{t_A - \frac{v}{c^2} x_A}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{t_O + \frac{L}{c} - \frac{v}{c^2} x_O - \frac{v}{c^2} L}{\sqrt{1 - \frac{v^2}{c^2}}} = t'_O + \frac{L}{c} \frac{ 1 - \frac{v}{c} }{\sqrt{1 - \frac{v^2}{c^2}}}
\\ x'_A = \frac{x_A - v t_A}{\sqrt{1 - \frac{v^2}{c^2}}} = x'_O + L \frac{1 - \frac{v}{c}}{\sqrt{1 - \frac{v^2}{c^2}}}
\\ t'_B = \frac{t_B - \frac{v}{c^2} x_B}{\sqrt{1 - \frac{v^2}{c^2}}} = t'_O + \frac{\frac{L}{u} - \frac{v}{c^2} L}{\sqrt{1 - \frac{v^2}{c^2}}} = t'_O + \frac{L}{u} \frac{1 - \frac{u v}{c^2} }{\sqrt{1 - \frac{v^2}{c^2}}}
\\ x'_B = \frac{x_B - v t_B}{\sqrt{1 - \frac{v^2}{c^2}}} = x'_O + \frac{ L - v \frac{L}{u}}{\sqrt{1 - \frac{v^2}{c^2}}} = x'_O + L \frac{ 1 - \frac{v}{u}}{\sqrt{1 - \frac{v^2}{c^2}}}
$$
This inverse Lorentz transform is just as general as the forward direction, so we may summarize by writing:
$$t' = \frac{t - \frac{v}{c^2} x}{\sqrt{1 - \frac{v^2}{c^2}}} \\ x' = \frac{x - v t}{\sqrt{1 - \frac{v^2}{c^2}}}$$
Step 2 Find the equations for a, b, d and e in terms of the new coordinates:
$$x_a - x_O = c ( t_a - t_O ) \Rightarrow \frac{x'_a + v t'_a}{\sqrt{1 - \frac{v^2}{c^2}}} - \frac{x'_O + v t'_O}{\sqrt{1 - \frac{v^2}{c^2}}} = c ( \frac{t'_a + \frac{v}{c^2} x'_a}{\sqrt{1 - \frac{v^2}{c^2}}} - \frac{t'_O + \frac{v}{c^2} x'_O}{\sqrt{1 - \frac{v^2}{c^2}}} )
\\ \Rightarrow x'_a + v t'_a - x'_O - v t'_O = c t'_a + \frac{v}{c} x'_a - c t'_O + \frac{v}{c} x'_O
\\ \Rightarrow \frac{1 - \frac{v}{c}}{\sqrt{1 - \frac{v^2}{c^2}}} (x'_a - x'_O ) = c \frac{1 - \frac{v}{c}}{\sqrt{1 - \frac{v^2}{c^2}}} ( t'_a - t'_O )
\\ \Rightarrow x'_a - x'_O = c ( t'_a - t'_O ) $$
Above we see that a has identical description with reference to event O in every coordinate system.
$$x_b - x_O = u ( t_b - t_O ) \Rightarrow \frac{x'_b + v t'_b}{\sqrt{1 - \frac{v^2}{c^2}}} - \frac{x'_O + v t'_O}{\sqrt{1 - \frac{v^2}{c^2}}} = u ( \frac{t'_b + \frac{v}{c^2} x'_b}{\sqrt{1 - \frac{v^2}{c^2}}} - \frac{t'_O + \frac{v}{c^2} x'_O}{\sqrt{1 - \frac{v^2}{c^2}}} )
\\ \Rightarrow x'_b - x'_O = \frac{u - v}{1 - \frac{u v}{c^2} } ( t'_b - t'_O ) $$
Above we see the origin of the Einstein velocity composition law. ($$u' = \frac{u + v}{1 + \frac{u v}{c^2} }$$ for the case where they are parallel and $$u' = \frac{u - v}{1 - \frac{u v}{c^2} }$$ (change of sign on v) for the case where u and v are oppositely oriented (antiparallel).)
$$x_d - x_O = 0 \Rightarrow x'_d - x'_O = (- v) ( t'_d - t'_O )
\\ x_e - x_O = L \Rightarrow x'_e - x'_O = \sqrt{1 - \frac{v^2}{c^2}} L - v ( t'_e - t'_O )$$
Now e and d don't correspond to a state of rest in these coordinates. (But since L was defined in the coordinate system where both d and e were at rest, above we see the origin of the Lorentz contraction of length, $$L' = \sqrt{1 - \frac{v^2}{c^2}} L $$.)
All four of these show that Newton's law of inertia is obeyed the same in both frames, so that tends to support Einstein's first postulate.
Step 3 Calculate $$C = a \cap b, D = a \cap e, E = b \cap e $$ in these new coordinates:
$$x'_C - x'_O = c ( t'_C - t'_O ), \quad x'_C - x'_O = \frac{u - v}{1 - \frac{u v}{c^2} } ( t'_C - t'_O )$$ with solution $$x'_C = x'_O, t'_C = t'_O$$.
$$x'_D - x'_O = c ( t'_D - t'_O ), \quad x'_D - x'_O = \sqrt{1 - \frac{v^2}{c^2}} L - v ( t'_D - t'_O )$$ with solution:
$$ t'_D = t'_O + \frac{\sqrt{1 - \frac{v^2}{c^2}}}{c + v} L = t'_O + \frac{L}{c} \frac{1 - \frac{v}{c}}{\sqrt{1 - \frac{v^2}{c^2}}} = t'_A
\\ x'_D = x'_O + L\frac{1 - \frac{v}{c}}{\sqrt{1 - \frac{v^2}{c^2}}} = x'_A$$
$$x'_E - x'_O = \frac{u - v}{1 - \frac{u v}{c^2} } ( t'_E - t'_O ), \quad x'_E - x'_O = \sqrt{1 - \frac{v^2}{c^2}} L - v ( t'_E - t'_O )$$ with solution:
$$ t'_E = t'_O + \frac{1 - \frac{u v}{c^2} }{\sqrt{1 - \frac{v^2}{c^2}} } \frac{L}{u} = t'_B
\\ x'_E = x'_O + \frac{u - v}{\sqrt{1 - \frac{v^2}{c^2}} } \frac{L}{u} = x'_B$$
Step 4 Highlight the logical self-contradiction.
But there is none since O and C, A and D, and B and E all identify the same event coordinates (respectively).
---
So in the OP the ratio $$r = \frac{t_A - t_O}{t_B - t_O} = \frac{u}{c}$$ is highlighted. In the transformed frame we defined $$r' = \frac{t'_A - t'_O}{t'_B - t'_O} = \frac{ c^2 u - c u v }{ c^3 - c u v } \neq \frac{u}{c}$$.
But that's not a violation of the first postulate because the law r = r' is not a law of physics. It's just a made-up ratio of coordinate times among three events which are related only pair-wise. You might as well be complaining that triangles have different heights when the coordinate system is rotated.
Here are quantities that SR does preserve:
$$ <A - O, A - O> = (c t_A - c t_O)(c t_A - c t_O) - (x_A - x_O)(x_A - x_O) = L^2 - L^2 = 0 \\ <A - O, A - O> = (c t'_A - c t'_O)(c t'_A - c t'_O) - (x'_A - x'_O)(x'_A - x'_O) = L^2 \frac{\left(1 - \frac{v}{c} \right)^2}{1 - \frac{v^2}{c^2}} - L^2 \frac{ \left( 1 - \frac{v}{c} \right)^2}{1 - \frac{v^2}{c^2}} = 0$$
Both coordinate systems agree, the trajectory from O to A is light-like.
$$ <B - O, B - O> = (c t_B - c t_O)(c t_B - c t_O) - (x_B - x_O)(x_B - x_O) =
\frac{c^2 L^2}{u^2} - L^2 = \frac{c^2 - u^2}{u^2} L^2
\\ <B - O, B - O> = (c t'_B - c t'_O)(c t'_B - c t'_O) - (x'_B - x'_O)(x'_B - x'_O) = \frac{\left(1 - \frac{u v}{c^2} \right)^2 }{1 - \frac{v^2}{c^2}} \frac{c^2 L^2}{u^2} - \frac{\left(u - v\right)^2}{1 - \frac{v^2}{c^2} } \frac{L^2}{u^2} = \frac{L^2}{u^2} \frac{c^2 - 2 u v + \frac{u^2 v^2}{c^2} - u^2 + 2 u v - v^2 }{1 - \frac{v^2}{c^2}} = \frac{c^2 - u^2}{u^2} L^2
$$
Both coordinate systems agree, the trajectory from O to B is time-like. The value is proportional to the square of the proper elapsed time.
$$ <A - O, B - O> = (c t_A - c t_O)(c t_B - c t_O) - (x_A - x_O)(x_B - x_O) = \frac{c^2 L^2}{c u} - L^2 = \frac{c - u}{u} L^2
\\ <A - O, B - O> = (c t'_A - c t'_O)(c t'_B - c t'_O) - (x'_A - x'_O)(x'_B - x'_O) =
L^2 \frac{c}{u} \frac{\left(1 - \frac{v}{c} \right)\left(1 - \frac{u v}{c^2} \right)}{1 - \frac{v^2}{c^2}} - L^2 \frac{ \left( 1 - \frac{v}{c} \right)\left(1 - \frac{v}{u}\right)}{1 - \frac{v^2}{c^2}} = L^2 \frac{ \frac{c}{u} - \frac{v}{u} - \frac{v}{c} + \frac{v^2}{c^2} - 1 + \frac{v}{c} + \frac{v}{u} - \frac{v^2}{c u} }{1 - \frac{v^2}{c^2}} = L^2 \left( \frac{c}{u} - 1 \right) = \frac{c - u}{u} L^2
$$
This is analogous to the dot product of Euclidean vectors.
---
Also in proper language, we can describe these events and trajectories in coordinates. In the inertial coordinates (x,t) where d and e are the standard of rest:
$$x_a - x_O = c ( t_a - t_O ) \\ x_b - x_O = u ( t_b - t_O ) \\ x_d - x_O = 0 \\ x_e - x_O = L $$
where L is the distance between d and e as measured in this frame, which makes much more sense for a name than "x".
To find the coordinates of $$A = a \cap e$$, we solve:
$$x_a - x_O = c ( t_a - t_O ) \\ x_e - x_O = L \\ x_A = x_a = x_e \\ t_A = t_a = t_e$$
with solution $$x_A = x_O + L , \; t_A = t_O + \frac{L}{c}$$
Likewise, the coordinates of $$B = b \cap e$$ are $$x_B = x_O + L , \; t_B = t_O + \frac{L}{u}$$
Now to check if the Lorentz equations are self-consistent, we need to transform a, b, d,and e into new coordinates and see if they give the same solution as directly transforming A and B. (We're using the language of algebra, as did Einstein in 1905, but in 1908 Minkowski demonstrated that one can also use the language of geometry which provides deeper math reasons for the basic self-consistency of Special Relativity.)
$$t = \frac{t' + \frac{v}{c^2} x'}{\sqrt{1 - \frac{v^2}{c^2}}} \\ x = \frac{x' + v t'}{\sqrt{1 - \frac{v^2}{c^2}}}$$
Step 1: Find the (original system) coordinates for O, A and B. I will illustrate the general solution with O:
$$t_O = \frac{t'_O + \frac{v}{c^2} x'_O}{\sqrt{1 - \frac{v^2}{c^2}}} \\ x_O = \frac{x'_O + v t'_O}{\sqrt{1 - \frac{v^2}{c^2}}}$$
Then
$$t_O - \frac{v}{c^2} x_O = \frac{t'_O + \frac{v}{c^2} x'_O - \frac{v}{c^2} x'_O - \frac{v}{c^2} v t'_O}{\sqrt{1 - \frac{v^2}{c^2}}} = t'_O \frac{1 - \frac{v^2}{c^2}}{\sqrt{1 - \frac{v^2}{c^2}}} = t'_O \sqrt{1 - \frac{v^2}{c^2}}$$
Thus $$t'_O = \frac{t_O - \frac{v}{c^2} x_O}{\sqrt{1 - \frac{v^2}{c^2}}}$$
Similarly: $$x'_O = \frac{x_O - v t_O}{\sqrt{1 - \frac{v^2}{c^2}}}$$
Likewise,
$$t'_A = \frac{t_A - \frac{v}{c^2} x_A}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{t_O + \frac{L}{c} - \frac{v}{c^2} x_O - \frac{v}{c^2} L}{\sqrt{1 - \frac{v^2}{c^2}}} = t'_O + \frac{L}{c} \frac{ 1 - \frac{v}{c} }{\sqrt{1 - \frac{v^2}{c^2}}}
\\ x'_A = \frac{x_A - v t_A}{\sqrt{1 - \frac{v^2}{c^2}}} = x'_O + L \frac{1 - \frac{v}{c}}{\sqrt{1 - \frac{v^2}{c^2}}}
\\ t'_B = \frac{t_B - \frac{v}{c^2} x_B}{\sqrt{1 - \frac{v^2}{c^2}}} = t'_O + \frac{\frac{L}{u} - \frac{v}{c^2} L}{\sqrt{1 - \frac{v^2}{c^2}}} = t'_O + \frac{L}{u} \frac{1 - \frac{u v}{c^2} }{\sqrt{1 - \frac{v^2}{c^2}}}
\\ x'_B = \frac{x_B - v t_B}{\sqrt{1 - \frac{v^2}{c^2}}} = x'_O + \frac{ L - v \frac{L}{u}}{\sqrt{1 - \frac{v^2}{c^2}}} = x'_O + L \frac{ 1 - \frac{v}{u}}{\sqrt{1 - \frac{v^2}{c^2}}}
$$
This inverse Lorentz transform is just as general as the forward direction, so we may summarize by writing:
$$t' = \frac{t - \frac{v}{c^2} x}{\sqrt{1 - \frac{v^2}{c^2}}} \\ x' = \frac{x - v t}{\sqrt{1 - \frac{v^2}{c^2}}}$$
Step 2 Find the equations for a, b, d and e in terms of the new coordinates:
$$x_a - x_O = c ( t_a - t_O ) \Rightarrow \frac{x'_a + v t'_a}{\sqrt{1 - \frac{v^2}{c^2}}} - \frac{x'_O + v t'_O}{\sqrt{1 - \frac{v^2}{c^2}}} = c ( \frac{t'_a + \frac{v}{c^2} x'_a}{\sqrt{1 - \frac{v^2}{c^2}}} - \frac{t'_O + \frac{v}{c^2} x'_O}{\sqrt{1 - \frac{v^2}{c^2}}} )
\\ \Rightarrow x'_a + v t'_a - x'_O - v t'_O = c t'_a + \frac{v}{c} x'_a - c t'_O + \frac{v}{c} x'_O
\\ \Rightarrow \frac{1 - \frac{v}{c}}{\sqrt{1 - \frac{v^2}{c^2}}} (x'_a - x'_O ) = c \frac{1 - \frac{v}{c}}{\sqrt{1 - \frac{v^2}{c^2}}} ( t'_a - t'_O )
\\ \Rightarrow x'_a - x'_O = c ( t'_a - t'_O ) $$
Above we see that a has identical description with reference to event O in every coordinate system.
$$x_b - x_O = u ( t_b - t_O ) \Rightarrow \frac{x'_b + v t'_b}{\sqrt{1 - \frac{v^2}{c^2}}} - \frac{x'_O + v t'_O}{\sqrt{1 - \frac{v^2}{c^2}}} = u ( \frac{t'_b + \frac{v}{c^2} x'_b}{\sqrt{1 - \frac{v^2}{c^2}}} - \frac{t'_O + \frac{v}{c^2} x'_O}{\sqrt{1 - \frac{v^2}{c^2}}} )
\\ \Rightarrow x'_b - x'_O = \frac{u - v}{1 - \frac{u v}{c^2} } ( t'_b - t'_O ) $$
Above we see the origin of the Einstein velocity composition law. ($$u' = \frac{u + v}{1 + \frac{u v}{c^2} }$$ for the case where they are parallel and $$u' = \frac{u - v}{1 - \frac{u v}{c^2} }$$ (change of sign on v) for the case where u and v are oppositely oriented (antiparallel).)
$$x_d - x_O = 0 \Rightarrow x'_d - x'_O = (- v) ( t'_d - t'_O )
\\ x_e - x_O = L \Rightarrow x'_e - x'_O = \sqrt{1 - \frac{v^2}{c^2}} L - v ( t'_e - t'_O )$$
Now e and d don't correspond to a state of rest in these coordinates. (But since L was defined in the coordinate system where both d and e were at rest, above we see the origin of the Lorentz contraction of length, $$L' = \sqrt{1 - \frac{v^2}{c^2}} L $$.)
All four of these show that Newton's law of inertia is obeyed the same in both frames, so that tends to support Einstein's first postulate.
Step 3 Calculate $$C = a \cap b, D = a \cap e, E = b \cap e $$ in these new coordinates:
$$x'_C - x'_O = c ( t'_C - t'_O ), \quad x'_C - x'_O = \frac{u - v}{1 - \frac{u v}{c^2} } ( t'_C - t'_O )$$ with solution $$x'_C = x'_O, t'_C = t'_O$$.
$$x'_D - x'_O = c ( t'_D - t'_O ), \quad x'_D - x'_O = \sqrt{1 - \frac{v^2}{c^2}} L - v ( t'_D - t'_O )$$ with solution:
$$ t'_D = t'_O + \frac{\sqrt{1 - \frac{v^2}{c^2}}}{c + v} L = t'_O + \frac{L}{c} \frac{1 - \frac{v}{c}}{\sqrt{1 - \frac{v^2}{c^2}}} = t'_A
\\ x'_D = x'_O + L\frac{1 - \frac{v}{c}}{\sqrt{1 - \frac{v^2}{c^2}}} = x'_A$$
$$x'_E - x'_O = \frac{u - v}{1 - \frac{u v}{c^2} } ( t'_E - t'_O ), \quad x'_E - x'_O = \sqrt{1 - \frac{v^2}{c^2}} L - v ( t'_E - t'_O )$$ with solution:
$$ t'_E = t'_O + \frac{1 - \frac{u v}{c^2} }{\sqrt{1 - \frac{v^2}{c^2}} } \frac{L}{u} = t'_B
\\ x'_E = x'_O + \frac{u - v}{\sqrt{1 - \frac{v^2}{c^2}} } \frac{L}{u} = x'_B$$
Step 4 Highlight the logical self-contradiction.
But there is none since O and C, A and D, and B and E all identify the same event coordinates (respectively).
---
So in the OP the ratio $$r = \frac{t_A - t_O}{t_B - t_O} = \frac{u}{c}$$ is highlighted. In the transformed frame we defined $$r' = \frac{t'_A - t'_O}{t'_B - t'_O} = \frac{ c^2 u - c u v }{ c^3 - c u v } \neq \frac{u}{c}$$.
But that's not a violation of the first postulate because the law r = r' is not a law of physics. It's just a made-up ratio of coordinate times among three events which are related only pair-wise. You might as well be complaining that triangles have different heights when the coordinate system is rotated.
Here are quantities that SR does preserve:
$$ <A - O, A - O> = (c t_A - c t_O)(c t_A - c t_O) - (x_A - x_O)(x_A - x_O) = L^2 - L^2 = 0 \\ <A - O, A - O> = (c t'_A - c t'_O)(c t'_A - c t'_O) - (x'_A - x'_O)(x'_A - x'_O) = L^2 \frac{\left(1 - \frac{v}{c} \right)^2}{1 - \frac{v^2}{c^2}} - L^2 \frac{ \left( 1 - \frac{v}{c} \right)^2}{1 - \frac{v^2}{c^2}} = 0$$
Both coordinate systems agree, the trajectory from O to A is light-like.
$$ <B - O, B - O> = (c t_B - c t_O)(c t_B - c t_O) - (x_B - x_O)(x_B - x_O) =
\frac{c^2 L^2}{u^2} - L^2 = \frac{c^2 - u^2}{u^2} L^2
\\ <B - O, B - O> = (c t'_B - c t'_O)(c t'_B - c t'_O) - (x'_B - x'_O)(x'_B - x'_O) = \frac{\left(1 - \frac{u v}{c^2} \right)^2 }{1 - \frac{v^2}{c^2}} \frac{c^2 L^2}{u^2} - \frac{\left(u - v\right)^2}{1 - \frac{v^2}{c^2} } \frac{L^2}{u^2} = \frac{L^2}{u^2} \frac{c^2 - 2 u v + \frac{u^2 v^2}{c^2} - u^2 + 2 u v - v^2 }{1 - \frac{v^2}{c^2}} = \frac{c^2 - u^2}{u^2} L^2
$$
Both coordinate systems agree, the trajectory from O to B is time-like. The value is proportional to the square of the proper elapsed time.
$$ <A - O, B - O> = (c t_A - c t_O)(c t_B - c t_O) - (x_A - x_O)(x_B - x_O) = \frac{c^2 L^2}{c u} - L^2 = \frac{c - u}{u} L^2
\\ <A - O, B - O> = (c t'_A - c t'_O)(c t'_B - c t'_O) - (x'_A - x'_O)(x'_B - x'_O) =
L^2 \frac{c}{u} \frac{\left(1 - \frac{v}{c} \right)\left(1 - \frac{u v}{c^2} \right)}{1 - \frac{v^2}{c^2}} - L^2 \frac{ \left( 1 - \frac{v}{c} \right)\left(1 - \frac{v}{u}\right)}{1 - \frac{v^2}{c^2}} = L^2 \frac{ \frac{c}{u} - \frac{v}{u} - \frac{v}{c} + \frac{v^2}{c^2} - 1 + \frac{v}{c} + \frac{v}{u} - \frac{v^2}{c u} }{1 - \frac{v^2}{c^2}} = L^2 \left( \frac{c}{u} - 1 \right) = \frac{c - u}{u} L^2
$$
This is analogous to the dot product of Euclidean vectors.
]