A zero kernel does not force monicity in a merely semiadditive category

For commutative monoids $M$ and $N$, the Cartesian product $M\times N$ is also their coproduct: if $f:M\to P$ and $g:N\to P$ are homomorphisms, then $h(m,n):=f(m)+g(n)$ is a homomorphism $M\times N\to P$, and it is the unique one with $h(m,0)=f(m)$ and $h(0,n)=g(n)$ because every $(m,n)$ equals $(m,0)+(0,n)$. Hence finite products and finite coproducts in $\mathbf{C}\mathbf{M}\mathbf{o}\mathbf{n}$ agree, so [L1] shows that $\mathbf{C}\mathbf{M}\mathbf{o}\mathbf{n}$ is semiadditive.

Let $s,t:\mathbb{N}\to \mathbb{N}$ be the monoid homomorphisms $s(n)=n$ and $t(n)=2n$. They are distinct, but $qs=qt$ because both send every positive integer to $a$ and $0$ to $0$. Hence $q$ is not monic.

The equalizer of $q$ and the zero map $\mathbb{N}\to \{0,a\}$ consists only of $0$, because $q(n)=0$ holds exactly for $n=0$. So the kernel of $q$ is zero.

Thus a zero kernel does not force monicity once one weakens preadditivity to mere semiadditivity.