I'm still confused by the question when $n>0$: $\pi_n^{\tau} X $ are already groups, so applying the group completion functor do not do anything to them, and, even in the catagory of spaces, I have never seen a general connection between the higher homotopy group of a monoid $M$ and its group completion... What type of result do you have in mind when $C=\{*\}$ and $\tau$ is trivial ?

But, anyway one do have (in space) that if $X$ is a "monoid objects" in space (either using Segal style definition, or $A_{\infty}$ monoids, or something else) the $\pi_0$ of the group completion of $X$ is indeed the universal group generated by the (ordinary) monoid $\pi_0(X)$, mostly because of the respective universal property of all the objects involved. And the same is true in any $\infty$-topos



Using $\infty$-categorical language, you have $C$ an $\infty$-category, and you consider some $\infty$-topos $Sh(C)$ which is a left exact localization of $pSh(C)$ (the $\infty$-category of presheaves of spaces on $C$).

The functor $\pi^{\tau}_0$ you describe is the left adjoint to inclusion: 

$$ Sh_{set}(C) \subset Sh(C) \subset pSh(C) $$

(where $Sh_{set}$ denotes the category of sheaves of sets, i.e. the $0$-truncated objects of $Sh(C)$ )

Group completion is left adjoint to the inclusion of "monoids objects" (which are simplicial object satisfying the segal condition and $X[0] \simeq 1$ into group objects (which further satisfies the special Segal condition).

One also easily see that the natural monoid/group structure on $\pi_0^{\tau}$ of a monoid/group object make it the left adjoint to the forgetfull functor:

$$ Monoid(Sh_{set}(C)) \subset Monoid(pSh(C)) $$

Now instead of thinking about commutation of left adjoint one can simply look at the commutation of right adjoint that are very simple forgetful functor.


If you have a group object in $Sh_{set} C$, and you either first forget it is a group (to only have a monoid) and then see it as a monoid object in $pSh(C)$, or if you first forget it is truncated (to see it as a group object in $Sh(C)$) and then forget the monoid structure you get the same objects. So as the right adjoint commutes, the corresponding left adjoint commutes as well.