Suppose we have a successor function $S\colon G\to G$ such that the recursive definition
$$\tag1 g\times e:=e,\qquad g\times S(h):=g\times h+g$$
makes sense.
Define a map $\phi\colon \Bbb N_0\to G$ by $\phi(0)=e$, $\phi(n+1)=S(\phi(n))$. Then $\phi$ must be onto (and in particular, $G$ must be countable) because otherwise $(1)$ would not define $\times$ completely.
By induction, we obtain
$$\tag{2}g\times \phi(n)=n\cdot g.$$

If $\phi$ is injective, it is bijective and we obtain an inverse bijection $\psi\colon G\to\Bbb N_0$.

If $\phi$ is not injective, there is a minimal $m$ with $f(m)=f(k)$ for some $k<m$. By surjectivity, it must be the case that $m=|G|$. Then as $g\times\phi(|G|)=|G|\cdot g=e$, we may as well adjust $S$ so that $\phi(|G|)=e$, i.e., (re-)define $S(\phi(|G|-1)):=e$. This allows us to view $\phi$ as map $\Bbb Z/|G|\Bbb Z\to G$, which is then a bijection and we get an inverser bijection $\psi\colon G\to \Bbb Z/|G|\Bbb Z$.

At any rate, we arrive at
$$\tag 3g\times h=\psi(g)\cdot g $$
for $\psi$ as defined above.

## Under what conditions will this be a semiring?

We have
$$ (a\times b)\times c=\psi(\psi(a)\cdot b)\cdot c$$
and
$$ a\times(b\times c)=\psi(a)\psi(b)\cdot c.$$
In general, these will differ. A *sufficient* condition for equality is that $\psi$ is additive, i.e., $(G,+)$ is isomorphic to one of $(\Bbb N_0,+)$, $(\Bbb Z/n\Bbb Z,+)$.

We have
$$ (a+b)\times c=\phi(c)\cdot(a+b)=\phi(c)\cdot a+\phi(c)\cdot b=a\times c+b\times c.$$
But
$$a\times (b+c)=\phi(b+c)\cdot a $$
will in general not be equal to
$$a\times b+a\times c=\phi(b)\cdot a+\phi(c)\cdot a=(\phi(b)+\phi(c))\cdot a.$$
Again, a sufficient condition would be that $\psi$ is an isomorphism.