I have a question about the following exercise: Suppose $I$ is an ideal (two-sided) in a semigroup $S$. Prove that $I$ is prime (which means that from $aSb\subset I$ for some $a,b\in S$ it follows that either $a$ or $b$ is in $I$) if and only if for any two ideals $A$ and $B$ in $S$ if $AB\subset I$ then either $A$ or $B$ is contained in $I$.
Here's my proof for necessity: take any two ideals $A,B\subset S$ with $AB\subset I$. Note that then $\cup_{a\in A, b\in B}aSb = ASB\subset AB\subset I$ (1). Suppose that neither $A$ nor $B$ are contained in $I$. In this case, there exist $a_0\in A\backslash I$ and $b_0\in B\backslash I$, which is clearly a contradiction, since $I$ is prime and due to (1) $a_0 S b_0\subset I$.
For sufficiency, I tried the following thing: take any $a,b\in S$ with $aSb\subset I$. Take $(SaS)(SbS)=SaSSbS\subset SaSbS \subset I$ (the latter inclusion is due to $I$ being an ideal and since $aSb\subset I$). Now according to our assumption, either $SaS\subset I$ or $SbS\subset I$ (they're both ideals, aren't they). Can I get from here the thing I want? Clearly, if $S$ has the identity element, the conclusion follows, but in case there's no identity, my argument seems to be a dead end.
It's not a homework assignment. Could anyone please check my proof of necessity and hint on sufficiency?