Can you reconcile your statement with an expression?:
$$(ct')^2-x'^2=(ct)^2-x^2$$
I already did:
The transit times are not supposed to be roots of $$(ct')^{2} \,-\, L^{2} \,=\, 0$$. That is not what the equation means or how you are supposed to apply it.
I can illustrate how this equation works in more detail, by considering more than one reference frame.
Start with the mirrors
at rest a distance
L from one another. Then $$\Delta x_{1} \,=\, -L$$ and $$\Delta x_{2} \,=\, +L$$, and $$\Delta t_{1} \,=\, -L/-c \,=\, L/c$$ and $$\Delta t_{2} \,=\, L/c$$.
Because the pulse is moving at the speed of light in both directions, $$(c \Delta t_{1})^{2} \,-\, (\Delta x_{1})^{2} \,=\, 0$$ and $$(c \Delta t_{2})^{2} \,-\, (\Delta x_{2})^{2} \,=\, 0$$.
You can transfer to the case where the mirrors are
moving with velocity
+v via a Lorentz transform:
$$
\begin{eqnarray}
\Delta t' &=& \gamma (\Delta t \,+\, \frac{v}{c^{2}} \Delta x) \\
\Delta x' &=& \gamma (\Delta x \,+\, v \Delta t) \,,
\end{eqnarray}
$$
with $$\gamma \,=\, \gamma(v) \,=\, \frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}$$.
You get different transit times in both directions because $$\Delta x_{1} \,=\, -\Delta x_{2}$$:
$$
\begin{eqnarray}
\Delta t'_{1} &=& \gamma (\Delta t_{1} \,+\, \frac{v}{c^{2}} \Delta x_{1}) \\
&=& \gamma (L/c \,-\, \frac{v}{c^{2}} L) \\
&=& (L/c) \gamma (1 \,-\, \frac{v}{c}) \\
&=& (L/c) \frac{1}{\gamma (1 \,+\, \frac{v}{c})} \\
&=& \frac{L'}{c \,+\, v} \\
\Delta t'_{2} &=& \gamma (\Delta t_{2} \,+\, \frac{v}{c^{2}} \Delta x_{2}) \\
&=& \gamma (L/c \,+\, \frac{v}{c^{2}} L) \\
&=& \frac{L'}{c \,-\, v} \,,
\end{eqnarray}
$$
with $$L' \,=\, L/\gamma$$, reflecting relativistic length contraction. To make the expressions more recognisable I used that $$\gamma \,=\, \frac{1}{\sqrt{1 \,+\, \frac{v}{c}} \sqrt{1 \,-\, \frac{v}{c}}}$$ to work out that $$\gamma (1 \,\mp\, \frac{v}{c}) \,=\, \frac{1}{\gamma (1 \,\pm\, \frac{v}{c})}$$.
So you get the different transit times like you expected: $$\Delta t'_{1} \,=\, L'/(c \,+\, v)$$ and $$\Delta t'_{2} \,=\, L'/(c \,-\, v)$$, and $$\Delta t'_{1} \,\neq\, \Delta t'_{2}$$.
According to Minkowski's equation,
$$
\begin{eqnarray}
(c \Delta t_{1})^{2} \,-\, (\Delta x_{1})^{2} \,=\, 0 &\Rightarrow& (c \Delta t'_{1})^{2} \,-\, (\Delta x'_{1})^{2} \,=\, 0 \,, \\
(c \Delta t_{2})^{2} \,-\, (\Delta x_{2})^{2} \,=\, 0 &\Rightarrow& (c \Delta t'_{2})^{2} \,-\, (\Delta x'_{2})^{2} \,=\, 0 \,.
\end{eqnarray}
$$
If you want to verify that this still holds, you need to work out $$\Delta x'_{1}$$ and $$\Delta x'_{2}$$. The transit distances are
not $$\pm L'$$, because the mirrors are moving while the pulse moves between them. You can work them out with the Lorentz transformation again:
$$
\begin{eqnarray}
\Delta x'_{1} &=& \gamma (\Delta x_{1} \,+\, \Delta t_{1}) \\
&=& \gamma (-L \,+\, v L/c) \\
&=& - L \gamma (1 \,+\, \frac{v}{c}) \\
&=& - \frac{L'}{1 \,+\, \frac{v}{c}} \\
\Delta x'_{2} &=& \gamma (\Delta x_{2} \,+\, \Delta t_{2}) \\
&=& \gamma (L \,+\, v L/c) \\
&=& \frac{L'}{1 \,-\, \frac{v}{c}} \,.
\end{eqnarray}
$$
Now you can see that $$\Delta x'_{1} \,=\, - \frac{L'}{1 \,+\, \frac{v}{c}} \,=\, - \frac{c L'}{c \,+\, v} \,=\, - c \Delta t'_{1}$$ and $$\Delta x'_{2} \,=\, c \Delta t'_{2}$$, so Minkowski's equation holds just fine:
$$
\begin{eqnarray}
(c \Delta t'_{1})^{2} \,-\, (\Delta x'_{1})^{2} \,=& (c \Delta t'_{1})^{2} \,-\, (- c \Delta t'_{1})^{2} &=\, 0 \,, \\
(c \Delta t'_{2})^{2} \,-\, (\Delta x'_{2})^{2} \,=& (c \Delta t'_{2})^{2} \,-\, (c \Delta t'_{2})^{2} &=\, 0 \,.
\end{eqnarray}
$$
Hopefully this rather long post will illustrate how the Lorentz transformation and Minkowski's equation apply to your problem.