Let $\Bbb M$ be a commutative semiring. Setting $b = 1 + 1$, $\, \Bbb M$ also satisfies
P-1: For every $k,h \in \Bbb M$ and $r \in \{0,1\}$
$$\tag 1 kb = hb + r \ \; \text{ iff } \; k = h \text{ and } r = 0$$
I'm trying to prove that $\Bbb M$ is infinite.
Set $k = 0$ and $h = 0$ and $r = 1$ and applying $\text{(1)}$,
$\quad 0b = 0b + 1$
is false. Since $0b = 0$, $\;1 \ne 0$.
Set $k = 0$ and $h = 1$ and $r = 0$ and applying $\text{(1)}$,
$\quad 0b = 1b + 0$
is false. But then $\;1+1 \ne 0$.
Set $k = 0$ and $h = 1$ and $r = 1$ and applying $\text{(1)}$,
$\quad 0b = 1b + 1$
is false. But then $\;1+1+1 \ne 0$.
I then convinced myself that the set
$$\{0,1,1+1,1+1+1,1+1+1+1,1+1+1+1+1\}$$
contains $6$ elements and using induction (not completely thought out) that the commutative semiring $\Bbb M$ is infinite.
Are there examples of these semirings that aren't isomorphic to the $\Bbb N$ or $\Bbb Z$?
If $\mathbb{M}$ is a finite semiring, then the elements $b^1,b^2...$ cannot all be distinct.
Case 1:
Suppose that we can choose $m,n\ge 1$ so that $b^m=b^n$ and $b^{m-1}\ne b^{n-1}$. Then $k=b^{n-1}, \ h=b^{m-1},\ r=0$ satisfies $kb=hb+r$.
Case 2:
Suppose that for any $m,n\ge 1$ with $b^m=b^n$ the equality $b^{m-1}=b^{n-1}$ also holds. Suppose furthermore that there exists an $m>0$ such that $b^m=b^0=1$ and $b^{m-1} \ne 1$. Then $h=b^{m-1}, \ k=1, \ r=1$ satisfies $kb=hb+r$.
Case 3:
Suppose that any $m,n\ge 1$ with $b^m=b^n$ satisfies $b^{m-1}=b^{n-1}$ and that for any $m>0$ such that $b^m=1$ we also have $b^{m-1}=1$. Then it can be shown that $b=1$. It follows that $h=0, k=1, r=1$ satisfies $kb=hb+r$.