Would "There is no greatest integer" be sufficient axiom to add to Peano to make the strengthened finite Ramsey theorem provable? If so, what example can be offered of an unprovable statement in the new theory? Instinct tells me there might not be one, but then I am quite naive.
What I like about this axiom is that it adds nothing, it only takes away, which is what any axiom that creates something that doesn't really exist must do.
On
"There is no greatest (non-negative) integer." trivially translates to the sentence "$\neg \exists m\ \forall n\ ( n \le m )$" over PA, where "$x \le y$" is short-form for "$\exists t\ ( x+t = y )$". This sentence is easily provable over PA, which I shall leave as an exercise.
Consequently, adding your proposed axiom to PA does nothing at all, and all unprovable sentences over PA remain unprovable over the new system. To even state Ramsey-type theorems in PA one has to go through some encoding, which is not a trivial matter, and it is inadvisable to attempt to understand that without a firm foundation in logic. I'm serious; running before being able to walk is dangerous. I shall yet again encourage you to spend a few months working through proper logic texts such as the first two listed in this post.
Presumably "strengthened finite Ramsey theorem" means the Paris-Harrington theorem; if not, what does it mean?
Peano arithmetic already proves "There is no largest natural number," so adding this statement to PA just yields PA again.
If you're interested in the axioms you need to add to (say) PA in order to prove certain arithmetic facts, you should look up reverse mathematics. But this is a fairly advanced subject, and you should be familiar with the basics of proof and model theory before you tackle it. Your question suggests you are not quite comfortable with the model theory of first-order logic, so I suggest you start there (Ebbinghaus-Flum-Thomas is a good source; Marker is better written and more advanced, but has typoes).
Meanwhile, (Rosser's strengthening of) Goedel's theorem constructively provides a sentence $\varphi$ which is undecidable in a theory $T$, whenever $T$ is computably axiomatizable, consistent, and contains (say) Robinson's $Q$. So I don't understand the last two sentences of your first paragraph.
EDIT: See also Why is the Axiom of Infinity necessary?.