Consider a cancellative monoid $S$ satisfying the left Ore condition, so it embeds in a group $G=S^{-1}S$. Consider also the integral monoid rings $\mathbb Z[S]$ and $\mathbb Z[G]$.


I have two, probably trivial, questions:


1. Can one prove that $S$ satisfies the left Ore condition (for rings) as a multiplicative set in $\mathbb Z[S]$? I have tried both to find references and to prove it myself, but without success.
2. Even if the answer to (1) is negative, one can still consider the map $\mathbb Z[S]\to \mathbb Z[G]$ and use it to make a left $\mathbb Z[S]$-module $M$ into a left $\mathbb Z[G]$-module
$$ 
\bar M:=\mathbb Z[G]\otimes\_{\mathbb Z[S]}M.
$$
Furthermore, there is a canonical map $\varphi \colon M\to \bar M$, defined by $\varphi(m):=1\otimes m$, for all $m\in M$. Is there an easy description for $\ker(\varphi)$? For example, is $\ker(\varphi)=\{m\in M:\exists s\in S, sm=0\}$?



Integral monoid rings and Ore conditions
Let $s\in S$ and $r\in \mathbb{Z}[S]$. We can then write $r=\sum\_{i=1}^{n}\alpha\_i t\_i$ for some $\alpha\_{i}\in \mathbb{Z}$ and some $t\_i\in S$.


The left Ore condition on $S$ implies that there exists some $u\_1$ and $v\_1$ (in $S$) with $u\_1t\_1 = v\_1 s$. Similarly, the left Ore condition gives us $u\_2$ and $v\_2$ with $u\_2(u\_1t\_2) = v\_2s$. Repeating, we see that $$(u\_n u\_{n-1}\cdots u\_2 u\_1)r\in \mathbb{Z}[S]s.$$ This shows that $S$ is a left Ore set inside the domain $\mathbb{Z}[S]$.


