FALSE: the Reidemeister-Schreier presentation needs no choice of transversal

Let $G=F(a,b)=⟨a,b\mid ⟩$, and let $H\le G$ be the subgroup of words with even exponent sum in $a$. The two right cosets are $H$ and $Ha$, so both $\{1,a\}$ and $\{1,a^{-1}\}$ are Schreier systems.

With the Schreier system $\{1,a\}$, the nontrivial Schreier generators are $b$, $a^2$, and $ab{a}^{-1}$. With the Schreier system $\{1,a^{-1}\}$, the corresponding nontrivial Schreier generators are $b$, $a^2$, and $a^{-1}ba$. These are different generator lists.

The subgroup presented is the same subgroup $H$, but the rewritten generators depend on the chosen representatives. Therefore the Reidemeister-Schreier presentation does depend on the transversal.