If the Grothendieck ring of a semiring on a free commutative monoid is unital, is the original semiring unital? Suppose $S$ is an associative semiring whose underling commutative monoid is free (in particular, cancellative) and that its Grothendieck ring $G(S)$ is a unital ring. Can we conclude that $S$ must be unital, and if not, is there a nice counter-example?
Alternatively, are there additional suppositions we can put on $S$ which would allow us to conclude the unitality of $S$ from that of $G(S)$?
The context is that I am struggling to disprove that a particular ring I have constructed is unital, and I have proven it is of the form $G(S)$ for $S$ which is not unital. I am hoping this helps me somehow.
Cross-posted on math.stackexchange.
 A: The answer is no.  Let $S$ be a finite meet semilattice without maximum.  For concreteness, take $S$ to be the proper subsets of $\{1,2\}$ under intersection. Let $\mathbb NS$ be the semigroup semiring of $S$.  Then the underlying additive monoid is free on $S$.  The Grothendieck ring is the semigroup ring $\mathbb ZS$ and this ring has an identity, but the identity cannot be expressed as a non-negative linear combination of elements of $S$.  One writes it using inclusion-exclusion (or Mobius inversion).
For my particular $S$, the proper subsets of $\{1,2\}$, the identity element of $\mathbb ZS$ is $e=\{1\}+\{2\}-\emptyset$, which does not belong to $\mathbb NS$.  Recalling that my semigroup multiplication is intersection, you can just check that $e$ multiplied by an element of $S$ returns that element.
For a more general meet semilattice $S$, the identity is $\sum_{s\in S}\sum_{t\leq s}t\mu(t,s)$ where $\mu$ is the Mobius function of $S$.
I think for general semirings there is no hope of a reasonable condition.  For example if $S$ is a finite left regular band (that is a semigroup satisfying the identities $xyx=xy$ and $x^2=x$), I showed with Margolis and Saliola that $\mathbb ZS$ has an identity if and only if certain posets associated to $S$ have connected Hasse diagrams.  But if $S$ is not a monoid, the identity will not belong to $\mathbb NS$. This is already a hard result. The meet semilattice case is just a commutative left regular band.
I also showed that for any finite inverse semigroup $S$, the semigroup ring $\mathbb ZS$ has an identity, but $\mathbb NS$ will not have an identity unless $S$ has an identity.  This also generalizes the meet semilattice case.
Update: Here are some more observations.  First, if $S$ is a semigroup, then the semigroup ring $\mathbb NS$ is unital if and only if $S$ is a monoid.  That is because you have the surjective augmentation homomorphism $\mathbb NS\to\mathbb N$ summing up the coefficients and since the only nonzero idempotent of $\mathbb N$ is $1$, we must have that the support of an identity element has one element of $S$ with coefficient $1$ and so $S$ is a monoid.
Second, let $S$ be a meet semilattice (possibly infinite) viewed as a semigroup under meet.  Then $\mathbb NS$ has an identity iff $S$ has a maximum.  But $\mathbb ZS$ is isomorphic to the locally constant functions with compact support from $\widehat{S}\to \mathbb Z$ under pointwise operations.  Here $\widehat{S}$ is the space of all non-zero semigroup homomorphisms $\phi\colon S\to \{0,1\}$ with the topology of pointwise convergence, where $\{0,1\}$ is a semigroup under usual multiplication.  Thus $\mathbb ZS$ has an identity if and only if $\widehat{S}$ is compact.  This is obviously a topological issue.
