My intuition is that this statement is false and here is my proof.
$\exists x(P(x) \to Q(x))$
$\exists x(\lnot P(x) \lor Q(x))$ using logical equivalence.
$\exists x\lnot P(x) \lor \exists x Q(x)$ using distributive properties of $\exists$ over $\lor$.
Assuming Q(x) is always false, we simply need 1 such x where $\lnot P(x)$ is true and that makes the entire statement true. The other expression can be transformed into this:
$\exists xP(x) \to \exists xQ(x)$
$\lnot\exists xP(x) \lor \exists xQ(x)$ using logical equivalence.
$\forall x\lnot P(x) \lor \exists xQ(x)$
If we assume Q(x) is always false, the only thing that makes this true is for all $\lnot P(x)$ to be true. For example, the domain could be all positive integers and $P(x) = x \le 10$. To be complete, let's assume $Q(x) = x \le -10$.
Therefore, the lefthand side of $\exists x(P(x) \to Q(x))$ is true and the righthand side of $(\exists xP(x) \to \exists xQ(x))$ is false.
I'm pretty sure I'm right but I would like another set of eyes to see if I did anything stupid.