I decided to expand upon my comment a bit. I'll denote the monoid in question by $M$ and assume that $M$ is cancellative and that the operation of $M$ is denoted additively (of course, if it isn't, then replace subtraction by reciprocal, etc...the notation for the monoid operation isn't terribly important).
So, going back to the case of $\mathbb{M}=\mathbb{N}_0:=\mathbb{N}\cup \{0\}$, what would we mean by something like -2? Assuming that we knew what subtraction meant, we could write -2 in many ways: $-2=2-4=5-7=131-133$, etc (just like there are many ways to write fractions: $\frac{1}{3}=\frac{3}{9}=\frac{7}{21}$, etc). This may smell a bit as though there's an equivalence relation afoot, and there is.
Consider the relation $\sim$ on $M\times M$ by $(a,b) \sim (c,d)\,\Leftrightarrow a+d=b+c$ (ie if we knew what subtraction was, we'd have $a-b=c-d$). It turns out that this is an equivalence relation (it would be a good exercise to provide a proof of this). We then define the quotient group $Q$ of $M$ to be the set of equivalence classes of $\sim$ on $M\times M$. The operation in $Q$, which I'll denote by $\oplus$ (so that we don't confuse it with the operation in $M$), is merely defined by
$[a,b]\oplus [c,d] = [a+c,b+d]$, where $[x,y]$ denotes the equivalence class of $(x,y)$ (ie $[x,y]=${$(a,b)\in M\times M\,\vert\,(a,b)\sim (x,y)$}).
So, we have $0_Q=[m,m]$ for any $m\in M$, and we can identify $m\in M$ with $[m,0]\in Q$. Of course, it's a worthwhile exercise to prove that the elements of $Q$ form a group under $\oplus$.