Let $R$ be a ring (without assuming identity or commutativity), and $P$ a proper ideal of $R$. Show that the following are equivalent:
(a) For ideals $A,B$: $AB\subseteq P$ implies $A\subseteq P$ or $B\subseteq P$.
(b) For right ideals $T,S$: $TS\subseteq P$ implies $T\subseteq P$ or $S\subseteq P$.
(c) For elements $a,b\in R$: $aRb\in P$ implies $a\in P$ or $b\in P$.
(b) $\implies $ (a) is trivial
(c) $\implies$ (b) is easy: Assume (c) and let $T,S$ be as in (b), with $T\not\subseteq P$. Fix $r\in T\backslash P$ and let $s\in S$. Then $rRs\in P$ since $T$ is a right ideal and by assumption $s\in P$.
I haven't been able to prove (a) $\implies $ (c) since my first idea was to realise $a,b$ as two sided ideals, however, the ideal generated by $a$ in a ring without unity is $\langle a\rangle =RaR + aR+Ra+\mathbb Z a$, which is a bit unwieldly. I am not convinced that multiplying the ideals $\langle a\rangle\cdot \langle b\rangle$ is assured to fall inside $P$ despite having $aRb\in P$. I am mostly concerned about the $a\cdot b$ terms, since there is no identity, it's not in $aRb$. Also the fact that each ideal $\langle a\rangle$ could well be the whole ring.