Let $G$ be a group and $F(G)$ the free group on $G$ as a set. There is a natural epimorphism $F(G)\to G$ that maps $[\sigma]\in F$ to $\sigma$, let $K$ be its kernel. Is the set $$X=\{[\sigma][\tau][\sigma\tau]^{-1}:\sigma,\tau\in G\}$$ a free basis of $K$?
Some proofs of the Schreier theorem (e.g. the one in Kargopolov) require that the Schreier transversal of $K$ have the identity of $F(G)$ represent $K$, but some don't seem to (e.g. the one in Rotman) and the transversal I use here doesn't either, so I wasn't sure if $X$ is a basis of $K$.
Yes it's a free basis. Certainly $\{[g] : g \ne e\} \cup \{e\}$ is a Schreier transversal. Applying Rotman, Theorem 11.47, carefully, a basis for the kernel is given by the nontrivial elements of the form $[a][x][ax]^{-1}$, with $a \ne e$, and $[e]$. This is the same as your basis.
If $|G| = n$ is finite, one can get away with just counting. Given that the elements $[x][y][xy]^{-1}$ are all duplicates when $x = e$, the proposed basis has cardinality $n(n-1)+1$. On the other hand the Schreier index formula implies that $K \cong F_{n(n-1)+1}$. Since the proposed basis certainly generates the kernel (insert easy argument), and since free groups are Hopfian, the proposed basis is indeed a free basis.