Counterexamples to the Grothendieck group of a commutative monoid respecting tensor products $\newcommand{\Q}{\mathbb{Q}}\newcommand{\N}{\mathbb{N}}\newcommand{\Z}{\mathbb{Z}}$Recall the definitions of the tensor product $\otimes_{\mathbb{N}}$ of commutative monoids (see also this note by Harold Simmons) and of the Grothendieck group $K_0$ of a commutative monoid (see also wikipedia). Given commutative monoids $A$ and $B$, we can apply the universal property of $K_0$ to the diagram

to get a morphism of abelian groups
$$K^{\otimes}_{0|A,B}\colon K_{0}(A\otimes_{\mathbb{N}}B)\to K_{0}(A)\otimes_{\mathbb{Z}}K_{0}(B).$$
Since we also have an isomorphism $K^{\otimes,1}_{0}\colon K_{0}(\mathbb{N})\overset{\sim}{\dashrightarrow}\mathbb{Z}$, these morphisms endow $K_{0}$ with the structure of an oplax monoidal functor
$$
                (K_{0},K^{\otimes}_{0},K^{\otimes,1}_{0})
                \colon
                (\mathsf{CMon},\otimes_{\mathbb{N}},\mathbb{N})
                \longrightarrow
                (\mathsf{Ab},\otimes_{\mathbb{Z}},\mathbb{Z}).
$$
(I think―I haven't actually checked the coherence conditions.)
Now, I'm quite confident that the map $K^\otimes_{0|A,B}$ is not always an isomorphism, so I'm trying to come up with a counterexample. Am I correct in thinking that $A=B=(\mathbb{Z},\cdot,1)$ is such an example?
On one side, we have $\mathbb{Z}\otimes_{\mathbb{N}}\mathbb{Z}\overset{\text{i think?}}{\cong}\Z\oplus\Z_{2}$, so
\begin{align*}
K_0(\Z\otimes_{\N}\Z) &\cong K_0(\Z\oplus\Z_2)\\
                      &\cong K_0(\Z)\oplus K_{0}(\Z_2)\\
                      &\cong \Q\oplus\Z_2,
\end{align*}
(where we've used the strong monoidality of $K_0$ with respect to direct sum), while on the other we have
\begin{align*}
K_0(\Z)\otimes_{\Z}K_0(\Z) &\cong \Q\otimes_{\Z}\Q\\
                           &\cong \Q
\end{align*}
(via here). Thus
$$K_0(\Z\otimes_{\N}\Z)\cong\Q\oplus\Z_2\neq\Q\cong K_0(\Z)\otimes_{\Z}K_0(\Z).$$

Question
Are the above calculations correct? Moreover, what are some other examples (assuming this is indeed one >_<) of commutative monoids $A$ and $B$ such that the map $K_{0}(A\otimes_{\mathbb{N}}B)\to K_{0}(A)\otimes_{\mathbb{Z}}K_{0}(B)$ is not an isomorphism?
 A: In fact, the canonical map $K_{0}(A\otimes_{\mathbb{N}}B)\to K_{0}(A)\otimes_{\mathbb{Z}}K_{0}(B)$ is always an isomorphism.  One way to see this is by observing that if $M$ is a commutative monoid then $K_0(M)$ is naturally isomorphic to $\mathbb{Z}\otimes_{\mathbb{N}} M$ (where here $\mathbb{Z}$ is the additive monoid).  You can prove this, for instance, by verifying that bilinear maps $\mu:\mathbb{Z}\times M\to N$ are in natural bijection with homomorphisms $f:M\to N$ that send each element of $M$ to an invertible element (by defining $\mu(a,m)=af(m)$ in one direction and defining $f(m)=\mu(1,m)$ in the other direction).
So, $$K_0(A\otimes_\mathbb{N} B)\cong \mathbb{Z}\otimes_\mathbb{N} A\otimes_{\mathbb{N}} B\cong (\mathbb{Z}\otimes_\mathbb{N}\mathbb{Z})\otimes_\mathbb{N} A\otimes_{\mathbb{N}} B\cong (\mathbb{Z}\otimes_\mathbb{N} A)\otimes_\mathbb{N}(\mathbb{Z}\otimes_\mathbb{N} B)\cong K_0(A)\otimes_\mathbb{N} K_0(B).$$  Since $\otimes_\mathbb{N}$ is the same as $\otimes_\mathbb{Z}$ when applied to commutative monoids that are actually already groups, this gives an isomorphism $K_{0}(A\otimes_{\mathbb{N}}B)\to K_{0}(A)\otimes_{\mathbb{Z}}K_{0}(B)$.  It is then a matter of definition-chasing to confirm that this isomorphism is the same as the map you are considering (for instance, by verifying that they both send elements of the form $[a\otimes b]$ (which generate the group $K_{0}(A\otimes_{\mathbb{N}}B)$) to $[a]\otimes [b]$)).
There are several errors in your proposed example.  I'm not sure how you got that $\mathbb{Z}\otimes_\mathbb{N}\mathbb{Z}\cong\mathbb{Z}\oplus\mathbb{Z}_2$; this does not seem correct since (for instance) the elements $p\otimes -1$ should be distinct invertible elements of $\mathbb{Z}\otimes_\mathbb{N}\mathbb{Z}$ for each prime $p$ whereas $\mathbb{Z}\oplus\mathbb{Z}_2$ has only finitely many invertible elements (assuming you mean the $\mathbb{Z}$ there to still be the multiplicative monoid).  Also, $K_0(\mathbb{Z})=0$, not $\mathbb{Q}$, since adjoining an inverse to $0$ kills everything ($0\cdot x=0$ implies $x=1$ if $0$ becomes invertible).
