Yes, under reasonable assumptions on the underlying measure space, the essentual supremum is again measurable.
To see this, I will assume that $(Y, B,\mu)$ is $\sigma$ finite. Assumptions on the other factor $(X, A)$ are not necessary. In general, some assumption like $\sigma$-finiteness is necessary, since by taking the counting measure, we can always achieve equality of esssup and the "ordinary" supremum and as you noted, this supremum need not be measurable.
Let us first assume that $f \geq 0$. Since $Y = \bigcup_n Y_n$ with $Y_n$ of finite measure and because of
$$
\Vert f(x, \cdot) \Vert_{L^\infty (Y)} = \sup_n \Vert f(x, \cdot) \Vert_{L^\infty (Y_n)}, \qquad \qquad (1)
$$
it suffices to assume that $\mu(Y) < \infty$.
Furthermore, by replacing $f$ by $f_N := f \cdot 1_{|f| \leq N}$ and noting
$$
\Vert f (x, \cdot) \Vert_\infty = \lim_N \Vert f_N (x, \cdot) \Vert_\infty, \qquad \qquad (2)
$$
we see that we can assume $f$ to be bounded. In particular (since $Y$ is of finite measure), we have $f(x, \cdot) \in L^p(\mu)$ for all $p$.
Now, this (Limit of $L^p$ norm) question shows
$$
\Vert f(x, \cdot) \Vert_\infty = \lim_n \Vert f(x, \cdot) \Vert_n,
$$
so that it suffices to show that $x \mapsto \Vert f(x, \cdot) \Vert_p$ is measurable for each $1 \leq p< \infty$. For this, it is enough to show that $x \mapsto \Vert f(x,\cdot) \Vert_p^p$ is measurable.
But $|f|^p$ is measurable with respect to the product $\sigma$ algebra and $Y$ is a finite measure space. Thus, part of the statement of the Fubini-Tonelli theorem implies that
$$
x \mapsto \int_Y |f(x,y)|^p \, d\mu(y) = \Vert f(x, \cdot)\Vert_p^p
$$
is measurable.
It remains to remove the assumption $f \geq 0$. As above, it suffices to assume that $f$ is bounded, say $|f| \leq M$. But then we have $M + f \geq 0$ and this function is measurable. Thus,
$$
x \mapsto \Vert M+f \Vert_\infty = \mathrm{esssup}_y \,\, M + f(x,y) = M + \mathrm{esssup}_y \,\, f(x,y)
$$
is measurable as seen above. By subtracting $M$ (which retains measurability), we obtain the claim.
EDIT: Here are some more details. First, let us prove (1) from above. For this, let $g = f(x, \cdot)$. It is trivial that $\Vert g \Vert_{L^\infty(Y_n)} \leq \Vert g \Vert_{L^\infty (Y)}$. For the converse direction, let $\alpha := \sup_n \Vert g \Vert_{L^\infty (Y_n)}$ and note $|g| \leq \alpha$ almost everywhere on $Y_n$ for each $n \in \Bbb{N}$. Since countable unions of null-sets are null-sets, we get $|g| \leq \alpha$ almost everywhere on $Y = \bigcup_n Y_n$ and thus $\Vert g \Vert_{L^\infty (Y_n)} \leq \alpha$, so that we have equality.
Now, note that if $(g_n)_n$ is a sequence of measurable functions, then $\sup_n g_n$ is also measurable, since we have $\sup_n g_n = \lim_k \max_{\ell = 1,\dots, k} g_k$ and it is easy to see that the maximum of finitely many measurable functions is again measurable.
The proof of equation (2) is similar to that of equation (1). Again, the estimate $\geq$ is trivial. Furthermore, if $\alpha := \lim_n \Vert f_N (x, \cdot) \Vert_\infty = \sup_n \Vert f_N (x, \cdot) \Vert_\infty$, then $|f_N| \leq \alpha$ almost everywhere for each $N$. Since $|f| = \lim_N |f_N|$, this implies $|f| \leq \alpha$ almost everywhere, so that we get equality.
