Given a (not necessarily commutative) ring $R$ and a monoid $M$, we can form the monoid ring $R[M]$ in the same way as we form the group ring. I would expect the monoid ring construction to be left adjoint to some "forgetful" functor, probably sending a ring to its underlying multiplicative monoid, but the universal property given in the wikipedia article suggests that it is not so simple, since it involves a pair of homomorphisms and a commutativity condition. This suggests to me that maybe it involves functors (into or out of) a coslice category (perhaps rings under R?).
What would be the correct way of stating this universal property as an adjunction? I am just curious.
Christian Sievers's comment has the right idea. Fix the ring $R$. An $R$-algebra is an $R$-bimodule, $M$, equipped with a map $\times:M\otimes M\rightarrow M$ which is associative and has an identity element. Note that $R[M]$ is not just a ring but actually an $R$-algebra.
Every $R$-algebra has an underlying multiplicative monoid given by forgetting about the $R$-module structure and only remembering $\times$. The functor $R[-]$ is left adjoint to this forgetful functor.
Abelian groups are exactly the same thing as $\mathbb Z$-modules. Similarly, rings are the same as $\mathbb Z$-algebras (because their underlying additive groups are $\mathbb Z$-modules, and their multiplication gives $\times$). Therefore the left adjoint to the forgetful functor which takes any ring and gives its multiplicative monoid is the functor $\mathbb Z[-]$.