If $e^2 = e$ in a ring $R$ [not necessarily commutative], then $Re$ is an ideal iff $eR(1-e) = 0$ I believe I have a proof for the forward direction. First, a little lemma: since $Re$ is an ideal, it is in particular a right ideal, and $1 \cdot e \in Re$ is right-absorptive w.r.t. $Re$, i.e. $er \in Re$ for all $r \in R$. Since $r$ is arbitrary here, $eR \subseteq Re$. To show $eR(1-e) = 0$, it suffices to show $er(1-e) = 0$ for all $r \in R$. By the lemma, we may write $er = r'e$ for some $r' \in R$, so $$er(1-e) = r'e(1-e) = r'(e-e^2) = r'(e-e) = 0$$ by the idempotency of $e$. I have yet to find a proof for the converse. $e^2 = e \implies re(1-e) = e(1-e)r = 0$, so the assumption $er(1-e) = 0$ demonstrates a weak commutativity, which was also used for the forward direction above, but I can't bring that to reveal that $re$ is right-absorptive w.r.t. $Re$ for all $r \in R$. Any hints? Thanks.

If $er(1-e)=0$ for every $r$, then $er=ere\in Re$.

Therefore $eR\subseteq Re$, and $Re$ absorbs $R$ on the right.