Just to confirm Jacob Lurie's comment above (**EDIT:** And the following has been corrected -- a previous version fell for a classic blunder as pointed out by Jacob Lurie below): the group completion of $S$ is $\Omega^\infty \tau_{\leq 1} \mathbb S$ as an infinite loop space. We can see this using a group completion lemma:

**Lemma:** (cf. [1]) Let $C$ be an $E_\infty$ space, and let $t \in \pi_0 C$. Then

The localized $E_\infty$ space $C[t^{-1}]$ agrees with the localization $t^{-1} C$ of $C$ with respect to $t$ as a $C$-module.

Moreover, let $C_\infty =
\varinjlim(C \xrightarrow t C \xrightarrow t \cdots)$. Then $t^{-1} C_\infty = t^{-1} C$.

Therefore $C_\infty = t^{-1} C$ if and only if $C_\infty$ is a $C[t^{-1}]$-module.

**Proof:**
(1) follows by the Yoneda lemma: on the category of $C[t^{-1}]$-modules, $t^{-1} C$ and $C[t^{-1}]$ both corepresent the forgetful functor to spaces. (2) holds because $t: t^{-1} C \to t^{-1}C$ is invertible. For (3), "only if" is obvious; "if" follows because $C[t^{-1}]$ modules are (by definition!) local with respect to the map $t: C \to C$ and hence with respect to transfinite composites thereof.

**Corollary:** The group completion $K(S)$ of $S$ is $\Omega^\infty \tau_{\leq 1} \mathbb S$.

**Proof:** Let $t: S \to S$ be the functor given by tensoring with $1$. Then in the notation of the lemma, $S_\infty$ is easily seen to have a similar description to $S$ but with objects $\mathbb Z$ instead of $\mathbb N$, and by inspection $t$ acts invertibly on $S_\infty$. So by the lemma, we have $K(S) = S[t^{-1}] = t^{-1}S = S_\infty$. This category looks a lot like $\Omega^\infty \tau_{\leq 1} \mathbb S$, and in fact we can see that they are the same because $\Omega^\infty \tau_{\leq 1} \mathbb S$ is a Picard 1-category, so the canonical functor from $K(\Sigma) = \Omega^\infty \mathbb S$ extends along $K(\Sigma) \to K(S)$. The extension is obviously a bijection on objects, and hits the involution on $1$ which generates the category symmetric monoidally so it is full as well. Since the hom-sets are finite, it is also faithful and thus an equivalence of categories.

**Corollary:** The universal functor $\Sigma \to K(\Sigma) = \Omega^\infty \mathbb S$ does not factor through $K(S)$, and hence does not factor through $S$.

**Proof:** If it did, that would be to say that $\tau_{\leq 1} \mathbb S$ splits off of $\mathbb S$, but it can't; for example $\eta^2 \neq 0$.

[1]: This version of the group completion theorem was based on Prop 6 in an expository note by Thomas Nikolaus, "The group completion theorem via localizations of ring spectra", Prop 6. The note is available from Nikolaus' website; here's a direct link which will directly download the pdf, <1 MB).