We can easily construct free $R$-module when $R$ is unital by setting $$R[S] = \{ f\colon S\to R\,|\, f\ \text{finitely supported}\}$$ and defining operations pointwise. The key here is that we can include the set of generators $S$ into $R[S]$ via map $s\mapsto \delta_s$ where $\delta_s(x) = 1_R$ if $x = s$ and zero otherwise. This allows us to uniquely express any $f\in R[S]$ as $$f=\sum_{s\in S} f(s)\delta_s$$ and thus extend by linearity any function $\varphi\colon S\to M$ to linear map $\overline\varphi\colon R[S]\to M$. Thus, $R[S]$ is indeed free.
Can we construct free module for non-unital rings? If possible, please give construction or reference.
P.S. One needs adequate notion of module here, of course. It's just representation of non-unital ring over abelian group, i.e. we just drop axiom $1_R\cdot a = a$.