Let $(S,+)$ a semigroup, is there exists a product over $S$, i.e., a binary operation $\cdot$ over $S$ such that $$ x\cdot(y+z)= x \cdot y + x \cdot z,\, (y+z)\cdot x = y\cdot x + z\cdot x,\, x\cdot(y\cdot z) = (x\cdot y)\cdot z$$ for all $x,y,z \in S$ ?
For example, for any finitely abelian group $G$, we know for the fundamental theorem of abelian groups that $G$ is isomorphic to a direct sum $\mathbb{Z}^n \oplus \mathbb{Z}_{p_1} \oplus \cdots \oplus \mathbb{Z}_{p_k}$, where $n,k \geq 0$ and $p_i$ is a prime tower. We can see each one of these groups as a ring under the usual product for $\mathbb{Z}$ and product mod $(p_i)$ for $\mathbb{Z}_{p_i}$, so this direct sum can be seen as a product ring, with $G$ as the underlying abelian group.
Assuming monoid $M$ is commutative with its operation denoted $+$, there is always the trivial product defined by $x \cdot y=0$ for all $x,y \in M$ [where $0$ is the additive identity of $M$ where the commutative monoid $M$ is written in additive notation.] This satisfies the rules you mention, but isn't a very interesting product to say the least. Note this construction works for any abelian monoid, finite or not.
Added later: As I noted in a comment one does not need to assume monoid $M$ is commutative. At the time I wasn't sure and asked for someone to check, which @J.-E. Pin has done.