Let $M,N$ be left $R$-modules with the standard $R$-module structures (making them bi-modules). The proof that $M \otimes N$ is isomorphic to $N \otimes M$ over a commutative ring $R$ in Dummit and Foote says that,
1) the map $(m,n) \mapsto n \otimes m$ is $R$-balanced and so induce a unique homomorphism $f(m \otimes n) = n \otimes m$.
2) Likewise, the map $(n,m) \mapsto m \otimes n$ is $R$-balanced and so induce a unique homomorphism $g(n \otimes m) = m \otimes n$.
Is it enough at this point to say that $f \circ g = $ id, and $g \circ f =$ id on the elementary tensors $n \otimes m$ and $m \otimes n$ respectively, and so they are isomorphisms? Or are there more to the proof that I am missing?
Yes it is!
The key here is to note that every element of $M \otimes N$ is a finite sum of elementary tensors $m \otimes n$.
However, not every element of $M \otimes N$ is an elementary tensor, but that's okay since $f$ linear implies $$\displaystyle f \bigg(\sum m_k \otimes n_k\bigg) = \sum f(m_k \otimes n_k).$$
This means it suffices to show that $g \circ f = \text{id}_{M \otimes N}$ and $f \circ g = \text{id}_{N \otimes M}$ on simple tensors (we just extend by linearity to the entire tensor product).