You are right: $X_t$ is not a Feller process. The following argument is somewhat sketchy, but it should not be too difficult to make it complete.

***

Let us consider the function $$u(x,y) = \tfrac{1}{8} (1 - x^{-2} - 4 y^2) $$ Observe that:

* the normal derivative of $u$ on the boundary of $D = \{(x, y) : |y| < e^{-x^4}\}$ is zero (except, of course, at $(0, 1)$ and at $(0, -1)$);

* $-\Delta u(x, y) = 1 + \tfrac{3}{4} x^{-4}$ is everywhere greater than $1$;

* $u(x, y) > 0$ if $(x, y) \in D$ and $|x| > 2$.

The above properties imply that $u(x, y)$ provides an upper bound for the mean hitting time $T_K$ of $K = \{(x, y) \in D : |x| \leqslant 2\}$ by $X_t$ (the reflected Brownian motion in $D$) started at $(x, y) \in D$, with $|x| > 2$.

Since $u$ is bounded above by $\tfrac{1}{8}$, the probability that $T_K \geqslant 1$ is at most $\tfrac{1}{4}$. It follows that with probability at least $\tfrac{3}{4}$, $T_K$ is less than one.

The above observation and the strong Markov property imply that $\mathbb{E}^{(x,y)}(\mathbb{1}_K(X_1))$ is bounded below by a positive constant in $D$, and so $X_t$ is not a Feller process.