Yes.

Using left modules, the indecomposable projective $eAe$-modules are $eP$ for $P$ an indecomposable projective $A$-module such that $e\operatorname{top}(P)=\operatorname{top}(P)$, and in this case $\operatorname{top}(eP)=\operatorname{top}(P)$.

For any finite-dimensional $A$-module $M$ such that $\operatorname{soc}(M)=e\operatorname{soc}(M)$, we have $\operatorname{soc}(eM)=\operatorname{soc}(M)$. Thus for the relevant indecomposable projectives we also have $\operatorname{soc}(eP)=\operatorname{soc}(P)$, and the result follows.