I write $[V,f]$ for the class of $(V,f)$ in your ring.

Let me prove the following property of $[V,f]$ : if $f: V\to W, g: W\to V$, then $[V,gf] = [W,fg]$.

Indeed, $[V,gf] = [V\oplus W, (gf, 0)]$, because $(W,0)$ is a commutator.

Further, by your trick about sums, $[P,-h] = -[P,h]$ so that $[V,gf] - [W,fg] = [V\oplus W, (gf,0)]+[V\oplus W, (0,-fg)] $ which, by using your trick again, is $[V\oplus W, (gf, -fg)]$. Now I claim that this is a commutator of the obvious things, namely $(v,w)\mapsto (0,f(v))$ and $(v,w)\mapsto (g(w),0)$.

This is a simple computation: applying the first composite to $(v,w)$ gives $(gf(v),0)$ and the second $(0,fg(w))$.

This proves the cyclic invariance claim. Now, by definition more or less, $HH_0(R)$ is the universal recipient of a cyclically invariant map $(V,f)\mapsto ``tr(V,f)"$. The trick here is to use cyclic invariance or summation with $(Q,0)$ to reduce from arbitrary projective to free, and then from free to ``free on 1 generator'' by using the standard basis of $End(R^n)$.

Conversely, sending $(V,f)$ to its Hattori-Stallings trace in $HH_0(R)$ satisfies your relation that commutators are sent to $0$.

So they are the same. For a commutative ring, $HH_0(R) = R$, as pointed out in the comments.

$\newcommand\HH{\mathit{HH}}$I write $[V,f]$ for the class of $(V,f)$ in your ring.

Let me prove the following property of $[V,f]$ : if $f: V\to W, g: W\to V$, then $[V,gf] = [W,fg]$.

Indeed, $[V,gf] = [V\oplus W, (gf, 0)]$, because $(W,0)$ is a commutator.

Further, by your trick about sums, $[P,-h] = -[P,h]$ so that $[V,gf] - [W,fg] = [V\oplus W, (gf,0)]+[V\oplus W, (0,-fg)] $ which, by using your trick again, is $[V\oplus W, (gf, -fg)]$. Now I claim that this is a commutator of the obvious things, namely $(v,w)\mapsto (0,f(v))$ and $(v,w)\mapsto (g(w),0)$.

This is a simple computation: applying the first composite to $(v,w)$ gives $(gf(v),0)$ and the second $(0,fg(w))$.

This proves the cyclic invariance claim. Now, by definition more or less, $\HH_0(R)$ is the universal recipient of a cyclically invariant map $(V,f)\mapsto “\operatorname{tr}(V,f)”$. The trick here is to use cyclic invariance or summation with $(Q,0)$ to reduce from arbitrary projective to free, and then from free to “free on 1 generator” by using the standard basis of $\operatorname{End}(R^n)$.

Conversely, sending $(V,f)$ to its Hattori–Stallings trace in $\HH_0(R)$ satisfies your relation that commutators are sent to $0$.

So they are the same. For a commutative ring, $\HH_0(R) = R$, as pointed out in the comments.