Let me start by interpreting the question "What kind of surface is $S$?" in the case of a general connected oriented topological surface (without boundary). (I am considering only oriented surfaces just for simplicity of discussion.) 
If $S$ had finite complexity, i.e., would be homeomorphic to the interior of a compact oriented surface, you probably would be satisfied by the answer of the type "$S$ is has $n$ ends and genus $g$", since this provides a complete set of topological invariants. Surfaces of infinite complexity are also classified by a certain set of invariants:

1. Its set of ends (regarded as a topological space).

2. Its genus. 

3. Its set of ends with positive genus.  

You can find more details and references in [this MO post](https://mathoverflow.net/questions/4155/classification-problem-for-non-compact-manifolds). 

If you look closely at the surface you are interested in, $H^2/[G,G]$, you realize that its invariants are:

1. The surface is 1-ended (simply because the abelian group $G/[G,G]$ is 1-ended). 

2. It has infinite genus (this is easy to see and is explained in Sam's answer). 

3. In particular, its only end has positive genus.  

To summarize: Your surface is the unique connected oriented topological surface of infinite genus and one end. If you are looking for a different answer, you should clarify what does your question really mean.