Let $S$ be an abelian group under an operation denoted by $+$. Suppose further that $S$ is closed under a commutative, associative law of multiplication denoted by $\cdot$. Say that $\cdot$ distributes over $+$ in the usual way. Finally, for every $s\in S$, suppose there exists some element $t$, not necessarily unique, such that $s\cdot t=s$.

Essentially, $S$ is one step removed from being a ring; the only problem is that the multiplicative identity is not unique. Here is an example.

Let $S=\{\text{Continuous functions} f: \mathbb{R}\rightarrow \mathbb{R} \ \text{with compact support}\}$ with addition and multiplication defined pointwise. It is clear that this is an abelian group with the necessary law of multiplication. Now, let $f\in S$ be supported on $[a,b]$. Let $S'\subset S$ be the set of continuous functions compactly supported on intervals containing $[a,b]$ that are identically 1 on $[a,b]$. Clearly, if $g\in S'$, then $f\cdot g=f$ for all $x$. Also, there is no unique multiplicative identity in this collection since the constant function 1 is not compactly supported.

I've observed that this example is an increasing union of rings, but I don't know if this holds for every set with the property I've defined.