Summary of what I have learned wrt this question. It turns out that the relevant construction is the *free covering semigroup* (first constructed by G. Cupona; thanks to W. Dudek for pointing this out to me), i.e. we have a functor $\mathsf{FreeCov}: \mathsf{nSgrp} \to \mathsf{Sgrp}$ which sends $\mathcal F_n(A) / \sigma \mapsto \mathcal F_2(A) / \sigma$ (as I suspected). We also have the *n-ary derivation* which is a functor $d_n: \mathsf{Sgrp} \to \mathsf{nSgrp}$. My question at hand (existence of an injection of any n-semigroup into a binary-derived n-semigroup) is satisfied by the evident injection
$$\mathcal F_n(A) / \sigma \hookrightarrow \mathcal d_n(\mathsf{FreeCov}(\mathcal F_n(A) / \sigma))$$

Also, it just so happens that $\mathsf{FreeCov}$ is left adjoint to $d_n$. Drawing from the pattern of $\mathsf{Free} \vdash \mathsf{Forgetful}$ we might wonder if $d_n$ is in some sense "forgetful". Yes. If we think of $\mathcal F_2(A) / \sigma$ as "knowing the value of" each string $a$, $ab$, $abc$, $abcd$, $abcde$ etc. (in the sense that the quotient by $\sigma$ defines a valuation map (of sets) $|\mathcal F_2(A)| \to A$, where $|-|:\mathsf{nSgrp} \to \mathsf{Set}$ is the underlying-set forgetful functor), then (fixing $n=4$ for example) $d_4(\mathcal F_2(A) / \sigma) = \mathcal F_4(A) / (\sigma \cap \mathcal F_4(A)^2)$ only "knows the value of" the strings with length $|s|\equiv 1 \bmod 3$, like $a$, $abcd$, $abcdefg$, etc., so it has "forgotten" the value of the strings of the other lengths.