Suppose $X_t = e^{-t}W_{e^{2t}}$ is an Ornstein–Uhlenbeck process ($W_t$ is a Wiener process). I'd like to show that for all $ 0 < a < b $,
$$ \mathbb{P}\Big( \max_{0 \leq t \leq a} X_t = \max_{0 \leq t \leq b} X_t\Big) = \frac{a}{b}. $$
I know I need to use the stationarity and Markov property of the Ornstein–Uhlenbeck process, but I'm not sure how to. Any help would be much appreciated.
I think you probably mean $X_t = e^{-t}W_{e^{2t}}$ instead. In either case, the important part is that $X_t$ is stationary and Markov, as you said. Put $$D_n = \left\{\frac{bk}{2^n} \mid k = 0, 1, \dots, 2^n\right\}$$ and $a_n$ the minimal member of $D_n$ which is at least $a$; this is exactly $a_n = \frac{b}{2^n} \cdot \lceil \frac{2^n a}{b} \rceil$. Note that $a_n \to a$ and is monotone decreasing.
Now consider $$P \left( \max_{0 \leq t \leq a_n} X_t = \max_{0 \leq t \leq b} X_t \right).$$ By continuity of measure, this sequence converges to the desired probability $$P \left( \max_{0 \leq t \leq a} X_t = \max_{0 \leq t \leq b} X_t \right),$$ but we know that by the fact that $X_t$ is stationary and Markov that $X_t$ on any interval $\left[\frac{k}{2^n}, \frac{k+1}{2^n}\right]$ has the same law as $X_t$ on any other such interval of the same length, and so the maximum of the process happens with uniform probability in any given one of those intervals, of which $0 \leq t \leq a_n$ contains $\lceil \frac{2^n a}{b} \rceil$. So $$ P \left( \max_{0 \leq t \leq a} X_t = \max_{0 \leq t \leq b} X_t \right) = \lim_{n \to \infty} P \left( \max_{0 \leq t \leq a_n} X_t = \max_{0 \leq t \leq b} X_t \right) = \lim_{n \to \infty} \frac{ \left\lceil \frac{2^n a}{b} \right\rceil}{2^n} = \frac{a}{b}. $$
Edit: I hand waved the calculation of the probability above. More precisely, we will do the following case for ease, though the general situation is not much harder. Let $X_t = e^{-t}W_{e^{2t}}$; we will show $$P\left(\max_{0 \leq t \leq 1/2} X_t \leq \max_{1/2 \leq t \leq 1} X_t\right) = \frac{1}{2}$$ explicitly. In particular, consider the time-reversed process $Y_t = X_{1-t} = e^{1-t}W_{e^{2 - 2t}}$. One can check that $E[Y_t] = 0$ and $\operatorname{Cov}(Y_t, Y_s) = e^{-|t-s|}$, so that $Y_t$ is also an OU process identical in law to $X_t$. Then, $$P\left(\max_{0 \leq t \leq 1/2} X_t \leq \max_{1/2 \leq t \leq 1} X_t\right) = P\left(\max_{0 \leq t \leq 1/2} Y_t \leq \max_{1/2 \leq t \leq 1} Y_t\right) = P\left(\max_{1/2 \leq t \leq 1} X_t \leq \max_{0 \leq t \leq 1/2} X_t\right)$$ so the maximum of $X_t$ on $[0,1]$ occurs with equal probability in $[0,1/2]$ and $[1/2, 1]$. The same logic extended gives the claim asked about in the comment.