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$$\forall y P(x) \implies \exists y Q(x,y)$$

Is it valid to rewrite the first part of the implication as $P(x)$ since the variable being quantified has nothing to do with the statement?

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Yes. If $y$ does not appear free in $P(x)$, then $\forall y P(x)$ and $P(x)$ are logically equivalent.

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    I have a question about this answer actually. If y does not appear free; but then the final result is re-used by another mathematician and y is substituted for x; will the result still be valid or will it not longer follow what was originally said? If not; does the the ∀y need to be carried through the working in some way so it appears in the final answer? – 2017-01-27
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    Whenever you do substitutions, you need to take the "usual measures" to ensure that bound variables don't get captured. The fine details of the "usual measures" vary in different presentations of first-order logic: one approach is to require all bound variables to have been renamed to avoid variable capture problems before the substitution is carried out. – 2017-01-27