When are semigroup-modules representable as rings?

Question

Let rings be commutative and unital.

I don't know what the standard terms are; I'm taking the notion of a $G$-module and generalizing it slightly. I might also be using the term representation in a weird way; I'm more than open to reworking the question to make the jargon (more) standard.

Suppose I have a semigroup $(S, \cdot)$ acting on an Abelian group $(A, +)$ via $\triangleright$. In other words, $\triangleright$ is a homomorphism of semigroups from $S$ to the endomorphism ring of $A$.

I will call this thing a semigroup-module.

What is a sufficient condition for a semigroup module having a ring representation? Some conditions that are necessary and sufficient would be ideal, but right now I just want some conditions that are collectively sufficient.

This notion comes from staring at a ring $R$ and asking what if we didn't know that elements used multiplicatively and elements used additively were the same?

Let's say that $(S, A, \triangleright)$ is representable if there exists a ring $R$ such that $(S, \cdot) \cong (R, \cdot)$ and $(A, +) \cong (R, +)$.

Let's suppose that $A$ has no elements such that $a+a = 0$ besides $0$, so I can include my condition on $-1$.

Being representable has some obvious consequences for the semigroup $S$:

1. It has an identity.
2. It has an annihilator.
3. It has the same cardinality as $A$.
4. It's commutative.
5. It has an involution that's not the identity.

However, I'm pretty sure that these conditions are not remotely sufficient for $(S, A, \triangleright)$ being representable.

To that end, I'm trying to come up with a semigroup module that's as simple as possible and passes 1-5 but isn't representable.

Here's what I have come up with as a candidate for a badly-behaved semigroup-module.

Let $S$ be $\{\pm 2^n : n \in \mathbb{N} \}$ with integer multiplication as our operation. Let $A$ be $(\mathbb{Z}, +)$. Let $S$ act on $A$ by multiplication.

$A$ and $S$ both satisfy the rules.

I think I can show that this semigroup-module is not representable using the following argument.

Suppose for contradiction there is such a ring $R$.

Let $1$ be the identity element of $S$.

By the fact that $\triangleright$ is an action, the element $1 + 1 + 1 \triangleright 1$ must be $1 + 1 + 1$.

However, there's nothing in $S$ which can send any $a$ to $a+a+a$. This is absurd, as was desired.

This makes it pretty clear that we need another necessary condition:

6. There exists a semigroup homomorphism $\varphi : (\mathbb{Z}, \cdot) \to S$ such that $\varphi(k) \triangleright a = \text{$a$ added to itself $k$ times}$.

I can keep playing condition whack-a-mole until I think I've found them all, but I'm curious whether there are any sufficient conditions for ring representations existing.