Let $M$ be a commutative monoid. We denote by $M^{Gr}$ its Grothendieck Group (i.e. group of fractions). We then have a morphism $q:M\rightarrow M^{Gr}$. We say that a commutative monoid $M$ is seminormal if $m\in M^{Gr}, m^2\in q(M)$ and $m^3\in q(M)$ implies that $m\in q(M)$. One also usually requires that $ab=ac$ in $M$ implies that $b=c$. Though I don't nescassarily require it.
Can somone give me an easy example of a monoid in generators and relations that is not seminormal but torsion-free (meaning $a^n=b^n$ implies that $a=b$).
Ii looks like $M = (\mathbb{N} - \{1\}, +)$ is a solution to your question. If I am right, the Grothendieck group of $M$ is $\mathbb{Z}$, and since $M$ is cancellative, $q$ is injective. Now $M$ is torsion-free, and taking $m = 1$, we get (in additive notation) $2m, 3m \in M$ but $m \in \mathbb{Z} - M$.
Since $\{2,3\}$ is a set of generators for $M$, a presentation for $M$ could be $$\langle a, b \mid 3a = 2b, a+b = b+a\rangle\ .$$