Hmm...

How about, since every group is a monoid, why not make \$$\mathbf{R}: \mathbf{Grp} \to \mathbf{Mon}\$$ simply the inclusion of groups into monoids, forgetting the inversion operation.

Also, I suppose if anything else, we could make a functor \$$\mathbf{Mon} \to \mathbf{Grp}\$$ by first taking any monoid and making a set from that monoid, \$$\mathbf{S} : \mathbf{Mon} \to \mathbf{Set}\$$; then taking that set and freely creating a group from that set, \$$\mathbf{T} : \mathbf{Set} \to \mathbf{Grp}\$$, since functors compose we get our desired functor, \$$\mathbf{L} = \mathbf{T}\circ\mathbf{S}\$$.

Then, I believe the following natural isomorphism is satisfied,

\$\mathbf{Grp}[L(m),g] \cong \mathbf{Mon}[m,R(g)] \$