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. So here is my question:

Q.Given an abelian group $G$, denote by $\mathscr B(G)$ themonoid 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 $\mathscr F(G) \to G$. For which $G$ is $\mathscr B(G)$ annular?

This is a very special case of a more general question (namely, when does a semigroup embeds into a ring etc.?), which, however, is also much harder and beyond the scope of this post. Indeed, my motivation is simply the following: Factorization theory (that is, the theory of non-unique factorization) grew up out of algebraic number theory, and a turning point in its history was when the theory, until then focused on integral domains, was reforged in the language of monoids, based at least in part on the consideration that the latter provide "models" (for studying various phenomena of interest) that ** wouldn't have been available otherwise**, with monoids of zero-sum sequences being probably the most representative of these models. So, the whole point of my question is that I'd like to understand to which extent the statement in bold is well-grounded.

*Update #1.* A fruitful exchange with Benjamin Steinberg in the comments below has eventually shown that if $H$ is a non-trivial monoid with trivial group of units (as in the case of interest here) embedding as a divisor-closed subsemigroup into the multiplicative monoid of a unital ring $R$, then the characteristic of $R$ is $2$, simply because the condition of divisor-closedness implies $R^\times \cong H^\times = \{1_H\}$.

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