Let $A$ be a (associative) ring with unity, $e$ an idempotent and $M$ a right $eAe$-module. I would like to work out this isomorphism $\mathrm{Hom}_{eAe}(A,N)=\mathrm{Hom}_{eAe}(Ae,N)$. It is related with question to which I give a link below:
$\mathrm{Hom}_{eAe}(A,N)=\mathrm{Hom}_{eAe}(Ae,N), N\otimes_{eAe}A=N\otimes_{eAe}eA$.
I know perfectly that $A=Ae\oplus A(1-e)$ as a right $eAe-$modules. It is obvious that $A(1-e)$ is annihilated by $eAe,$ but I don't know exactly how to define the isomorphism given above. If I map a homomorphism $f:A\rightarrow N$ to the restriction to $Ae,$ I have problem to prove the injectivity of this map. Is it possible that homomorphism $f,g:A\rightarrow N$ are equal on $Ae,$ but different out $Ae$? Annihilation stated above tells us only that image of element $A(1-e)$ under $f$ is annihilated by $eAe,$ not that it is zero... I suppose that $f(A(1-e))=0$ for any homomorphism $f:A\rightarrow N$ but I don't know if it is generally true and how to prove that.
For the question to make sense, you need that $A$ is a right $eAe$-module; in particular, it should be unital, where the unit of $eAe$ is of course $e$. So, you need that $ae=a$ for all $a\in A$, but this clearly only holds for the summand $Ae$. So, $A$ is not a right $eAe$-module.
In general, the idempotent $e$ determines a functor $$ F \colon \mathrm{mod}\,A \to \mathrm{mod}\,eAe, \quad X\mapsto Xe. $$ We have $Xe\cong X\otimes_AAe\cong\mathrm{Hom}_A(eA,X)$, so our functor $F$ has a right adjoint $M\mapsto\mathrm{Hom}_{eAe}(Ae,M)$ and a left adjoint $M\mapsto M\otimes_{eAe}eA$.