In general, if $f:R\to S$ is a surjective ring homomorphism, then the map Spec $S\to$ Spec $R$ is a closed embedding (in particular, it's injective).

Edit: I missed that $e$ may not necessarily be in $A$; the following assumes that it is.

If $e$ is an idempotent then so is $1-e$, and $A$ decomposes as a product $A = Ae \times A(1-e)$.  From the perspective of schemes, this product turns into a coproduct, and Spec $A$ is the disjoint union of Spec $Ae$ and Spec $A(1-e)$.  The schemes of the form Spec $Ae$ are hence those obtained as a union of some subset of the connected components of Spec $A$.