Here is the context for my question: Given a ring $R$ and monoid $M$, $R[M]$ is the set of maps $\varphi : M \to R$ where only finitely many elements do not map to the zero of $R$. This can be equipped with addition $$\varphi + \psi : m \mapsto \phi(m) + \psi(m)$$ and multiplication $$\varphi\psi : m \mapsto \displaystyle\sum_{l, n \in M, \text{ s.t. } ln = m} \varphi(l) \psi(n).$$
Does this definition of multiplication simplify at all if $R$ is commutative and $M$ is a commutative monoid?
Furthermore, what happens if $R = \mathbb{Z}_2$ and $M$ is commutative and idempotent?
My thought for the latter case is that $\varphi\psi : m \mapsto \varphi(m)\psi(m)$.
If $M$ is commutative, then $ln=m$ if and only if $nl=m$ for any $l,n,m\in M$. Since $R$ is also commutative, we have $\phi\psi(m)=\sum_{ln=m}\phi(l)\psi(n)=\sum_{ln=m}\psi(n)\phi(l)=\sum_{nl=m}\psi(n)\phi(l)=\psi\phi(m)$, so $R[M]$ is commutative, too.
If $R=\mathbb{Z}_{2}$ and $M$ is a commutative, idempotent monoid, then $R[M]$ is commutative and idempotent, too. Commutativity we have already shown. Let $\leq$ be an arbitrary total ordering on $M$. Then for any $\phi$ and any $m$ in $M$ we have \begin{align}\phi \phi(m)&=\sum_{l<n,ln=m}(\phi(l)\phi(n)+\phi(n)\phi(l))+\sum_{ll=m}\phi(l)\phi(l)\\ &=\sum_{l<n,ln=m}2\phi(l)\phi(n)+\sum_{ll=m}\phi(l)^{2}.\end{align} Since $R=\mathbb{Z}_{2}$, all terms in the first sum are $0$ and $\phi(l)^{2}=\phi(l)$. Since $M$ is idempotent, the only $l$ with $ll=m$ is $m$ itself, so we have shown $\phi\phi(m)=\phi(m)$.
Maybe one can say even more, but I don't see what or how at the moment.