There's a remarkable theorem of Margulis that pertains to your question. Let $G$ be a semisimple Lie group (in your case, $PSL_2(\mathbb{R})$), and let $\Gamma$ be an irreducible lattice in $G$. The commensurator of $\Gamma$ is defined as $$Comm(\Gamma)=\{ g\in G\ |\ [\Gamma: g\Gamma g^{-1}\cap \Gamma]<\infty\},$$ the subset of $G$ which "almost" normalizes $\Gamma$. For example, if $\Gamma <\Gamma'$ a larger lattice, then $\Gamma' \leq Comm(\Gamma)$ (notice that if $\Gamma$ is not normal in $\Gamma'$, then $g\Gamma g^{-1}\neq \Gamma$ for some $g\in \Gamma'$).
Margulis' theorem implies that either $Comm(\Gamma)$ is discrete, or $\Gamma$ is arithmetic. If $\Gamma$ is non-uniform (and $G=PSL_2(\mathbb{R})$), so $\mathbb{H}^2/\Gamma$ is finite volume but non-compact, then $\Gamma$ is commensurable with $PSL_2(\mathbb{Z})$, in other words, there is $g\in PSL_2(\mathbb{R})$ so that $[\Gamma: \Gamma \cap g PSL_2(\mathbb{Z}) g^{-1}] <\infty$. If $\Gamma$ is uniform, so $\mathbb{H}^2/\Gamma$ is a compact Riemann surface, then the description is a bit more complicated, but basically $\Gamma$ is commensurable with a discrete group defined by integral automorphisms of a ternary quadratic form of signature $(2,1)$ defined over a totally real number field, and which is definite at all other infinite places, using the isogeny $PGL_2(\mathbb{R})\cong O(2,1;\mathbb{R})$ (I think in the number theory lingo these are known as Shimura curves).
Now, suppose in your question that $X$ and $Y$ are finite area (in the unique complete hyperbolic metric given by the uniformization theorem; I can't say much of anything about the infinite area case), so define finite-sheeted covers of $E$. We may associate a torsion-free lattice $\Gamma < PSL_2(\mathbb{R})$ so that $E=\mathbb{H}^2/\Gamma$, and finite-index subgroups $\chi, \gamma < \Gamma$ so that $\mathbb{H}^2/\chi = X, \mathbb{H}^2/\gamma=Y$. Suppose $X$ and $Y$ are isomorphic Riemann surfaces, then there is $g\in PSL_2(\mathbb{R})$ so that $\chi = g\gamma g^{-1}$. Therefore $\Gamma > \chi = g \gamma g^{-1} < g \Gamma g^{-1}$, so $[\Gamma : \Gamma \cap g\Gamma g^{-1}] <\infty$, so $g\in Comm(\Gamma)$.
Thus, by Margulis' theorem, either $\Gamma$ is arithmetic, or $\Gamma$ is non-arithmetic, and therefore $\Lambda=Comm(\Gamma)$ is discrete. Then $\mathbb{H}^2/\Lambda$ is a finite-area hyperbolic orbifold $\mathcal{O}_\Lambda$, and one has that $g\Lambda g^{-1} =\Lambda$. This implies that $E\to \mathcal{O}_\Lambda$ is a finite-sheeted orbifold cover, and the two covers $X,Y\to E\to \mathcal{O}_\Lambda$ are equivalent (irregular) covers of $\mathcal{O}_\Lambda$. Notice that if $\mathcal{O}_\Lambda$ has non-trivial modulus (so it's not a turnover), then this gives rise to parameter space of such covers of Riemann surfaces of the same genus as $E$.
In the case $\Gamma$ is arithmetic, $Comm(\Gamma)$ is much larger. For example, $Comm(PSL_2(\mathbb{Z}))\cong PGL_2(\mathbb{Q})$. Such pairs of covers then are an important area of study in the theory of autormorphic forms, since they give rise to Hecke operators. However, for a fixed genus, there are only finitely many arithmetic Riemann surfaces of that genus (see e.g. Long-Reid for a more general result for orbifolds), so in particular there are only finitely many such examples for which $X$ and $Y$ have bounded area. There is (in principal) a complete arithmetic prescription for how such covers may occur in this case as well which is determined by the description of the commensurator.