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 we have a similar result for any $\infty$-topos (so any topology on any simplicial category).



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.

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Another way to rephrase this is that if $M$ is a monoid and $G$ is its group completion, then $\pi_0^{\tau} G$ and the (sheafification of) the group completion of $\pi_0^{\tau} M$ are equivalent because they have the same universal property:

Any morphism from $M$ to a sheaf of (discrete, i.e. $0$-truncated) group on $C$ factors "essentially uniquely" though the map $M \rightarrow \pi_0^{\tau} G$, where essentially uniquely means up to a contractible space of choices.

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Note that however it might not be the case that if you do the group completion of $\pi_0^{\tau} M$ as a presheaf of monoid, the result is sheaf ! You need to take the sheafification of this presheaf of group to get the correct "group completion" that has the universal properties explained above (but here we are talking about sheafification of ordinary presheaf of sets, with no homotopy theory involved).