The free-abelian-group monad sends a set to its finite formal integer combinations

By [L1]–[L4], the free-$\mathbb{Z}$-module on $X$ is ${\mathbb{Z}}^{(X)}$. For a function $u:X\to A$ into an abelian group, the unique extension sends ${\sum }_x{n}_x[x]$ to ${\sum }_x{n}_{x}u(x)$, so this is also exactly the universal property of the free abelian group on $X$.

Its unit is $x\mapsto 1[x]$. Its multiplication flattens ${\sum }_i{n}_i[{\sum }_j{m}_{ij}[x_{ij}]]$ to ${\sum }_{i,j}{n}_i{m}_{ij}[x_{ij}]$.

Every $\mathbb{Z}$-module is an abelian group under addition. Conversely, on an abelian group define $n\cdot x$ by repeated addition for positive $n$, by $0$ for $n=0$, and by negatives for negative $n$; the abelian-group laws give the module laws, and group homomorphisms are exactly the resulting $\mathbb{Z}$-linear maps. Hence the algebras are precisely abelian groups.