Let $A \subset \Bbb{Z}$ be any set closed under taking $\gcd$, then $\{a\Bbb{Z} + u : a \in A, u \in U\}$, where $U$ is any multiplicative subgroup of $\{-1, 1\}$, forms an elementwise monoid under usual ring multiplication because:
$$ (a\Bbb{Z} + u)(b\Bbb{Z} + v) = ab\Bbb{Z}\Bbb{Z} +av\Bbb{Z} + bu\Bbb{Z} + uv $$
but $\Bbb{Z}$ cancel's (absorbs?) units so:
$$ \gcd(a,b)\Bbb{Z} + uv $$
is the law, and since $\gcd$ is associative, we have a monoid in which $0\Bbb{Z}$ + 1 is identity.
My question is: Is the Grothendieck group of this commutative monoid trivial?
My next question is: Can we somehow form a group using a similar construction and elementwise operations on some subset of the cosets, that you know of, and the example must be interesting i.e. at least not trivial!
Feel free to extend $U$ to any subset of $\Bbb{Z}$ if need be!
Also, yes, I'm already aware of the modular arithmetic groups when you fix $a$ and take $U = \{ u \in \Bbb{Z}: \gcd(u,a) = 1\}$, i.e. the group of units modulo $a$.
We always have $(a\mathbb{Z}+1)^2=a\mathbb{Z}+1$, so this leaves only the cosets of the form $a\mathbb{Z}-1$ to deal with. Meanwhile, by the same argument any two cosets with "constant term" $-1$ wind up equal. So the Grothendieck group is just $\mathbb{Z}/2\mathbb{Z}$.