If I understand correctly, you say that an $R$-module $M$ is free if there exists a subset $X$ of $M$ such that the map$$ R^{(X)}\to M: (r_x)_{x\in X}\mapsto \sum_{x\in X}r_x\cdot x$$is bijective.
In that case, the following pretty much answers all your questions:If $R$ is commutative and if there exists a free $R$-module, then $R$ has a unity.
Indeed, assume that $M$ is a free $R$-module with basis $X$. Fix $x_0\in X$; there exists a family of elements of $R$ with finite support $(r_x)_{x\in X}$ such that $$x_0 = \sum_{x\in X}r_x\cdot x.$$
For any non-zero $s\in R$, $sx_0$ is non-zero, so $$- sx_0 + \sum_{x\in X}sr_x\cdot x = 0.$$
By the injectivity of the map above, we get $sr_x=0$ whenever $x\neq x_0$, and $sr_{x_0}-s=0$. Thus, for all $s\in R$, we have that $sr_{x_0} = s$; in other words, $r_{x_0}$ is a unity in $R$.