(Warning: the examples that I give here are all quite trivial compared to the motivating example.)
In general the group completion of a commutative monoid can have a much simpler structure than the monoid itself. An extreme example is the case of a monoid $M$ with an absorbing element, i.e., an element $z$ with $z*x = x*z = z$ for all $x \in M$. Then the group completion will just be the trivial group.
There are natural examples of monoids with absorbing elements. For instance, on p.5
I give the example of the commutative monoid of cardinalities of at most countable sets. This is the usual natural numbers together with an additional (absorbing) point at infinity. It has the natural structure of a semiring, so it is somewhat disappointing that its ring completion is trivial.
More generally, if you take a submonoid $M$ (in particular, a subset!) of the cardinal numbers under addition such that $M$ contains infinite cardinals, then you need not have an absorbing element but nevertheless the group completion will be trivial.
This essentially amounts to the example of a totally ordered set $(X,\leq)$ with a least element $0$ made into a commutative monoid via $x+y = \max(x,y)$.
Addendum: As Yemon Choi has pointed out below, a yet weaker condition for a commutative monoid to have trivial group completion is that $x+x = x$ for all $x \in M$. There are very rich classes of monoids satisfying this property!