There's a standard way to define the trace (look up "Hattori-Stallings trace") that agrees with yours, but is clearly independent of choices.

For any (left) $A$-module $P$, there's a natural map 
$$\operatorname{Hom}_A(P,A)\otimes_AP\to\operatorname{End}_A(P),$$
sending $\varphi\otimes y$ to the endomorphism $x\mapsto\varphi(x)y$, which is an isomorphism when $P$ is finitely generated projective. Composing its inverse with the map
$$\operatorname{Hom}_A(P,A)\otimes_AP\to A/[A,A]$$
induced by the evaluation map
$$\operatorname{Hom}_A(P,A)\otimes_kP\to A,$$
gives the trace map 
$$\operatorname{Tr}_P:\operatorname{End}_A(P)\to A/[A,A].$$