Ring with no identity (that has a subring with identity) has zero divisors. Let $L$ be a non-trivial subring with identity of a ring $R$. Prove that if $R$ has no identity, then $R$ has zero divisors.
So I assumed that there $\exists$ $e \in L$, such that $ex=xe=x$, $\forall$ $x\in L$ and there $\exists$ $x' \in R/L$ such that $ex'\neq x'$ and $x'e\neq x'$. I then show (we obtain it from the last two inequalities), that we have $y=ey=ye,$ where $y=ex', y\neq x'$, which is a contradiction because $y\notin L.$ Is there a way to show from this that $R$ has zero divisors? Because in one of the steps I have $e(x'-y)=0$, but I can't seem to figure out if this can directly lead to the desired result.

Actually you can prove something even better: a nonzero idempotent in a rng without nonzero zero divisors is actually an identity for the rng.. This applies to your case since the identity of the subrng would be a nonzero idempotent.
The linked solution proves the contrapositive of your statement very simply, and along analogous lines to yours.
