$\DeclareMathOperator\End{End}\newcommand\Set{\mathrm{Set}}\DeclareMathOperator\Nerve{Nerve}\DeclareMathOperator\Fun{Fun}$Here's perhaps a way to understand the general case that Neil mentioned, and a way to apply it to your question at the end.

Suppose you have a category $C$. Then there's a free symmetric monoidal category on $C$, $FC$, which Neil described in his answer.

Now if you're looking at $F^0(C,\Set)$ with respect to some $c\in C$, the category that you get only depends on $B{\End(c)}\subset C$, the full subcategory on $c$ (which is a one-object category, so you can see it as a monoid somehow), and in fact its core-groupoid (which is the only thing that $K$-theory depends on) is exactly $F(B{\End(c)^\times})$, i.e. finite free $\End(c)^\times$-sets, where $\End(c)^\times$ is the subgroup of invertible elements of $\End(c)$.

The reason for this is that left Kan extension along $\{c\}\to C$ factors as left Kan extension along $\{c\}\to B{\End(c)}$, and then left Kan extension along $B{\End(c)}\to C$, but the latter is a full subcategory inclusion, therefore left Kan extension along it is one as well, so if you're looking at the category of people that are left Kan extended from $\{c\}$ (from a finite set I would assume, to avoid Eilenberg swindle type phenomena), you might as well look at the same subcategory, but in $\Fun(B{\End(c)},\Set)$, so you might as well assume $C = B{\End(c)}$. Then you can notice that a good description of this category is just finite copies of $\End(c)$ with its $\End(c)$-action, and the core-groupoid of that will just be the same thing but for $\End(c)^\times$.

Therefore you get exactly the same situation as for a group : $K(\Fun^0(C,\Set)) = \Sigma^\infty_+(B(\End(c)^\times))$ where my $B$ is your $\lvert\Nerve({-})\rvert$.

Now the reason for that thing is that when $C$ is a groupoid (e.g. $B(\End(c)^\times)$), (the nerve of) $FC$ happens to also be the free symmetric monoidal $\infty$-groupoid on $C$, i.e. the free $E_\infty$-space on $\lvert\Nerve(C)\rvert$; hence if you apply group completion to it, you get the free grouplike $E_\infty$-space on it, in other words (up to delooping) the free connective spectrum on it. But that is precisely $\Sigma^\infty_+\lvert\Nerve(C)\rvert$. As Neil pointed out, this gets you the tom Dieck splitting, and in fact with the right setup for $G$-spectra it can give you an "equivariant Barratt–Priddy–Quillen" theorem.

Certainly this idea is present in Segal's *Categories and cohomology theories* — where he gives a proof "along those lines" of the Barratt–Priddy–Quillen theorem. A modern reference where this idea is explicitly used that way to prove the tom Dieck splitting is Barwick's *Spectral Mackey functors and equivariant algebraic $K$-theory (I)*, specifically theorem A.9. His argument can be generalized to describe the free $E_\infty$-space on a $1$-groupoid.