If a surface, M, has $\pi_1(M)=\mathbb{Z}\ast…\ast \mathbb{Z}$, is M a finitely punctured closed surface?

Yes. An open surface is either a finitely punctured surface or one has infinitely generated homology groups. A very light sketch is to pick an exhaustion by compact connected submanifolds; at each stage you're either gluing on cylinders, capping off boundary components with discs (or in the nonorientable case Mobius bands), or increasing the genus. If your homology is finitely generated, then your genus is finite, so eventually you're only adding on punctured spheres or capping off components; then show that adding punctured spheres without capping stuff off increases the rank of the homology, bounding the number of ends you can have.

A more precise argument in the case of $\pi_1 = 0$ is given here. You can modify the argument for the case of finitely-generated $H_1(\Sigma;\Bbb Z/2)$.

There is a classification of open surfaces following similar ideas: they are classified by their genus, whether or not they're orientable, and some data about precisely how they're noncompact: their space of ends, their space of non-orientable ends, and their space of ends with genus. If the fundamental group is finitely generated, it can't have any non-orientable ends or ends with genus; these contribute an infinitely generated summand to first homology. (This is why it's nice to work with homology instead of fundamental groups: Having an infinitely generated subgroup is ok, but not if you're abelian!)

At this point, we know that our surface is of finite genus, but it still might have a complicated space of ends (each end being genus zero). Each end contributes to homology: a manifold with at least $n$ ends has homology of rank at least $n-1$. So you had better have finitely many. Thus, you're a finitely punctured closed surface.