Let $K$ be a multiplicatively written semigroup (either commutative or not) and $H$ a subsemigroup of $K$. We say that $H$ is divisor-closed (in $K$) if $x \in H$ for all $x, y \in K$ such that $x \mid_K y$ (i.e., $y = uxv$ for some $u, v \in K$) and $y \in H$.

Accordingly, we say that a semigroup $S$ is annular (bear with me, I don't have a better word for the moment) if it embeds as a divisor-closed subsemigroup into the multiplicative monoid of a ring (either commutative or not). So here is my question:

What is known, if anything at all, about the abelian groups $G$ (either finite or not) for which $\mathscr B(G)$ is annular? Here, $\mathscr B(G)$ denotes the monoid of zero-sum sequences over $G$, that is, the submonoid of $\mathscr F(G)$, the free abelian monoid with basis $G$, given by the inverse image of $0_G$ under the canonical (monoid) epimorphism $\mathscr F(G) \to G$.

This is a very special case of a more general question (namely, when does a semigroup embeds into a ring etc.?), which, however, I'm afraid is also much harder.