B.L. van der Waerden "Modern Algebra, Volume II" (that volume was published in 1931) clearly identifies the notion of a free module (of finite rank), without using the term. I cite (English edition, p.98), where module means right K-module over a not necessarily commutative ring K:
The module $\mathfrak M$ is said to be finite (over K) if its elements may be represented linearly in the form $ u_1\lambda_1+\cdots+u_n\lambda_n $ by means of a finite number of elements $u_1,\ldots,u_n$. In this case we write $ \mathfrak M = (u_1\mathrm K,\ldots,u_n\mathrm K)\hbox{ or } \mathfrak M = (u_1,\ldots,u_n). $ If we further assume that the $u_i$ are linearly independent, that is $\sum u_i\alpha_i=0$ implies $\alpha_i=0$, then $\mathfrak M$ is called a ($n$-termed) module of linear forms or an $n$-dimensional vector space (cf. Section 14).
So clearly the notion is known by then, and the notion of vector space is even considered to encompass this rather general case. Given the quest for generality (he doesn't initially even assume that the identity of K acts as the identity on the module, though he quickly shows a reduction to that case!) I don't see why finite rank is assumed here from the start.