Commutative monoid, unsure of how to deal with negative elements, inverses and subtraction We are working with a commutative monoid.  Subtraction might be useful for us.
However, we're not sure how to proceed -- negative elements have no meaning.


*

*How do we deal with allowing subtraction in certain cases, but not in others?

*What algebraic structure (if any) do we apply here?

I think our problem is similar to this:
The commutative monoid is the non-negative integers, together with addition.  4 - 2 = 2 would be useful, but 2 - 4 = ??? would not be useful, since the answer would not be a non-negative integer.
 A: The result of the subtraction $a - b$ is any element $c$, if it exists, such that $a = b + c$. Unfortunately, in a general commutative monoid,


*

*$c$ may not exist, or

*there may exist more than one such $c$.


The non-negative integers gives a simple example of the first case. As a simple example of the second case, consider the non-negative integers under the operation $\text{min}(a, b)$. 
In other words, in general subtraction is partially defined and multivalued; it is better thought of as a relation than as a function.
I am not sure what "working with" means here so I don't know what else would be useful to say. 

To explain Jack Maney's comment, associated to any commutative monoid $M$ is a universal map to an abelian group $M \to A$. The group $A$ is called the Grothendieck group of $M$ and is, roughly speaking, the minimal way to adjoin inverses to $M$. It can explicitly be constructed as the quotient of the monoid of pairs $(m, n) \in M^2$ under componentwise addition by the following equivalence relation:
$$(m, n) \sim (p, q) \Leftrightarrow \exists r : m + q + r = n + p + r.$$
(One should think of $(m, n)$ as formally representing $m - n$.) When fed the non-negative integers this construction returns the integers. Unfortunately, the map $M \to A$ is not injective in general, so one loses information in general. It is injective if and only if $M$ is cancellative.
A: 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 $\oplus$.  
