From Wikipedia: An object $P$ in a category $\mathcal{C}$ is projective if for any epimorphism $e:E\twoheadrightarrow X$ and morphism $f:P\to X$, there is a morphism $\overline{f}:P\to E$ such that $e\circ \overline{f}=f$, i.e. the following diagram commutes:
Now, for reasons I find hard to convey, I would find the above definition more intuitive if $f$ were also an epimorphism.
My question is, over an abelian category, does restricting the above definition so that $f$ is also an epimorphism make the definition weaker?
They are equivalent. In fact, it suffices to consider the case of $f = 1_P$, which is an isomorphism and hence epi.
The special case of $1_P$ states that for all epis $e : E \to P$, there is some $m : P \to E$ such that $e \circ m = 1_P$.
To show this is sufficient, consider the diagram you have drawn with $P, E, X, e, f$. Take a pullback $D$ with morphisms $p_1 : D \to P$ and $p_2 : D \to E$.
In an Abelian category (and in many other kinds of categories), the pullback of an epi is epi, so $p_1: D \to P$ is epi. Take $m : P \to D$ such that $p_1 \circ m = 1_P$. Then $f = f \circ p_1 \circ m = e \circ p_2 \circ m$, so we can take $\tilde{f} = p_2 \circ m$.