This question only concerns the final part of the proof, so I assume that the symmetric monoidal category is a strict monoidal category $\mathsf{C}$ with the braiding $s$.

Let $X_1,...,X_n$ be elements of $\mathsf{C}$ and let $\sigma \in S_n$. Mac Lane proves that any two morphisms $f\colon X_1\otimes ... \otimes X_n \to X_{\sigma(1)}\otimes ... \otimes X_{\sigma(n)}$ which are compositions of braidings possibly tensored with identity morphism on the left or on the right $k$-times are equal. To this end, he realizes every such path a composition $s^{\pm}_{X_{i_1}, X_{i_1 + 1}} \circ ... \circ s^{\pm}_{X_{i_m}, X_{i_m + 1}}$ (I ignore $-\otimes 1_X$ and $1_X\otimes -$ here as far as the notation goes). The key part of the proof in an apparently intereseting connection between $s_{X_i,X_i + 1}$ and $(i,i+1)$.

Mac Lane states and any "closed path" (I suppose he's refering to said morphism with codomain $X_1\otimes ... \otimes X_n$) corresponds to a relations between generators $(i,i+1)$ of the symmetric group $S_n$. On the other hand, he mentions that a symmetric group has the presentation $$\langle \tau_i, i = 1,...n-1 \mid \tau^2_i = 1, (\tau_i\tau_{i+1})^3 = 1, \tau_i\tau_j = \tau_j\tau_i \text{ for }|i - j| > 1 \}$$

where $\tau_i = (i,i+1)$. He then claims that to prove the statement it suffices to show that these relations hold for $s$.

I've been thinking for a while about this and I can't understand what is the precise connection between braidings and permutations. I see that applying a braiding $s_{X_i,X_{i + 1}}$ to $X_1\otimes ... \otimes X_n$ given b$X_{\sigma(1)} \otimes ... \otimes X_{\sigma(n)}$ where $\sigma = (i,i+1)$, but no better than that yet.

Also, connecting relations in $S_n$ with those among $s_{X,Y}$ and deducing the statement of the theorem from that reminds me of the universal property of a presentation:

Let $\langle X \mid R \rangle$ be a presentation and $G$ a group. Let $f\colon X\to G$ be a map such that for every $(u,v) \in R$ we have $f(u) = f(v)$. Then there is a unique group homomorphism $\phi\colon\langle X \mid R \rangle \to G$ such that $\phi(x) = f(x)$ for all $x \in X$.

But I don't understand how this can be applied here as I don't see a resonable binary operation for paths consisting of braidings.

So what Mac Lane really means here, and how it can be made precise?