Consider a finite semigroup $S$ whose semigroup operation $\times$ is commutative and whose elements are idempotent.
Does there exist a finite union-closed family of finite sets $\mathcal{M}$ such that there is a bijection $\tau: S \rightarrow \mathcal{M}$ where $\forall a,b \in S, \tau(a \times b) = \tau(a) \cup \tau(b) $?
I feel that this may not be the case necessarily since I haven't encoded any information about "subsets,containment..." into the axioms i've stipulated on the semi group, but from a purely algebraic standpoint it would seem that this characterizes the union very well.
Yes. Given $a\in S$, consider the set $I(a)=\{b\in S:ab=b\}$. I claim that that $I(ab)=I(a)\cap I(b)$. Indeed, if $c\in I(a)\cap I(b)$ then $(ab)c=a(bc)=ac=c$ so $c\in I(ab)$. Conversely, if $c\in I(ab)$, then $ac=a(abc)=a^2bc=abc=c$ so $c\in I(a)$, and similarly $c\in I(b)$. Also, if $I(a)=I(b)$, then $a\in I(b)$ and $b\in I(a)$ so $a=ab=b$.
So, this is almost what you wanted, but with intersections instead of unions. To get unions, you can just take complements and define $\tau(a)=S\setminus I(a)$ and let $\mathcal{M}$ be the image of $\tau$.
Semigroups of this sort are known as (unbounded) semilattices and are a basic object of study in order theory. They are often thought of as ordered sets via the ordering $a\leq b$ if $ab=b$ (or $ab=a$, depending on the context); the algebraic structure can then be defined in terms of the ordering.