Here's another way to think about it.
If you want to prove the statement $(\exists x)(Ax \to Mx)$, then you need to find an $x$ such that $Ax \to Mx$ is true. Well, if $Ax$ and $Mx$ are both false (for instance, $x$ is a woman from another country - say, Angela Merkel), then $Ax \to Mx$ is true.
But have you really proven that there is a man in America by pointing out that Angela Merkel is a German woman?
No. So something is wrong with your formulation.
Note, that as a rule, $(\forall x)$ usually has $\to$ in it, and $(\exists x)$ usually has an $\wedge$ in it.