Question: Homological algebra for abelian groups is a standard tool in many fields of mathematics. How much carries over to the setting of commutative monoids (with unit)? It seems like there is a notion of short exact sequence. Can we use this to define ext groups which classify extensions? What works and what doesn't work and why?

Answer: 
This is an answer to one part of your question. The paper “Extension Theories for Monoids” by Charles Wells, Semigroup Forum 16 (1978), 13-35, gives a precise answer to the specific question: How does the Beck cohomology theory for monoids classify extensions of monoids? (It classifies Leech extensions.) The paper with corrections and a list of subsequent papers related to it may be found [here](http://www.cwru.edu/artsci/math/wells/pub/pdf/ExtThMon.pdf). Beck's thesis is now online [here](http://www.tac.mta.ca/tac/reprints/articles/2/tr2.pdf).
I agree with Charles and Eric that the natural setting for your question is the model category of simplicial commutative monoids. However, my earlier guess that the resulting homotopy theory would admit a simple algebraic description (as in the case of simplicial abelian groups) was wrong. For instance, consider the free commutative monoid functor from simplicial sets to simplicial commutative monoids and apply it to a space $X$ which I will assume for convenience to be connected and reduced ($X$ has only a single 0-simplex). The resulting simplicial commutative monoid $Y$ has $\pi\_0(Y) = \mathbb{N}$. The element $1 \in \mathbb{N}$ is distinguished as the unique generator, and the connected component of $Y$ corresponding to $1 \in \pi\_0(Y)$ is just $X$. So, we can recover $X$ from $Y$. 

In other words, when Charles says "A non-connected topological commutative monoid (or equivalently, a non-connected simplicial commutative monoid) contains some more information than its group completion", it is actually quite a lot more information, enough to encode an entire homotopy type!

If we apply the reduced chain complex functor to one of these free simplicial commutative monoids, we'll just get $\mathbb{N}$ in degree 0 and zero in higher degree. The problem is that knowing the kernel of a map is 0 tells us very little (*every* map in the image of the free commutative monoid functor has kernel 0).