Let $S$ be a semigroup and $I,J$ be two ideals of $S$. For a semilattice we know that $IJ=I\cap J$. Now the question is there a semigroup with the property $IJ=I\cap J$. thanks for your attention
I assume you are asking which semigroups have the property that the intersection of any two ideals is their product.
One condition that will do it is von Neumann regularity. $S$ is regular if, for all $s\in S$, there is $t\in S$ with $sts=s$.
In any semigroup $IJ\subseteq I\cap J$. If $S$ is regular and $s\in I\cap J$, then choosing $t$ with $sts=s$, we have $s\in I$ and $ts\in J$ and so $s=sts\in IJ$. Thus $IJ=I\cap J$.
Steinberg's example is more related to semigroup theory. But why you do not use ring theory? For every two comaximal ideals in a ring with one, we have this property. Note that
Every ideal of a ring, is an ideal of multiplicative semigroup of the ring
Every monoid which is the multiplicative monoid of a ring can be used for creating your example.
example Let S be the multiplicative monoid of integers.
suppose I=2Z and J=3Z