$\def\QQ{\mathbb Q}$If $\pi_1(M)$ is finite, $H_1(M;\QQ)=0$. If $M$ is non-orientable, $H_3(M;\QQ)=0$. So $\chi(M)=h_0(M;\QQ)-h_1(M;\QQ)+h_2(M;\QQ)-h_3(M;\QQ)=1+h_2(M;\QQ)>0$.
But by Poincaré duality, any odd-dimensional manifold has zero Euler characteristic.
If $M$ is a connected closed (i.e. compact without boundary) 3-dimensional manifold, then $\pi_1(M)$ cannot be isomorphic to $\pi_1(S)$, where $S$ is an orientable surface of genus $\ge 2$. You can see this by first noting that $\pi_2(M)=0$ (otherwise, by the sphere theorem, $M$ is a nontrivial connected sum which will imply that $\pi_1(M)$ is a nontrivial free product, which is not the case). Then you observe that $\pi_k(M)=0, k\ge 3$ since the universal cover $\tilde M$ of $M$ is noncompact (here you use that $S$ has positive genus, which implies that the fundamental group is infinite): Thus, $H_k(M)=0$ for all $k\ge 2$. Now, use Hurewicz theorem to conclude that $\pi_k(M)=0$ for all $k\ge 3$.
Thus, $M$ would be homotopy equivalent to $S$ (they have isomorphic fundamental groups and contractible universal covers; now, use Whitehead's theorem). However, $H_3(M)\ne 0$ while $H_3(S)=0$. Contradiction.
What one may ask is about topological classification of compact orientable 3-dimensional manifolds such that $\pi_1(M)\cong \pi_1(S)$. You can still have manifolds of the form: $S\times I$ minus some disjoint open balls. However, such manifolds which are reducible, meaning that there exists a tame 2-sphere in $M$ which does not bound a balls.
It is a theorem (I think, due to Stallings), that if, $M$ is assumed to be oriented, $S$ is oriented, and $M$ is irreducible then, it indeed is homeomorphic to $S\times I$. You can find a proof of this in Hempel's book "3-manifolds". The key is that in this situation there exists a proper homotopy-equivalence $h: M\to S\times I$, i.e., a homotopy equivalence which is a homeomorphism on the boundary. One then proves that such $h$ is homotopic (rel. boundary) to a homeomorphism.
If one allows for connected boundary of $M$, or for nonorientable $M$ or $S$, then the conclusion is that $M$ is homeomorphic to an $I$-bundle over $S$.
Edit. Here is a construction of an open 3-manifold which is homotopy-equivalent to a closed surface $S$ but is not homeomorphic to an interval bundle over $S$. Start with an open contractible 3-dimensional manifold $W$ which is not homeomorphic to $R^3$, say, the Whitehead manifold. Embed (properly and smoothly) a ray $\rho$ into $W$ and then remove a small open tubular neighborhood of $\rho$ from $W$. The result is a contractible manifold with boundary $X$; $\partial X$ is homeomorphic to the open disk $D^2$. Now, take $M=S\times I$, pick an disk $D\subset \partial M$ and glue $M$ to $X$ identifying $\partial X$ and $D$ homeomorphically. Lastly, remove the remaining boundary from the result of gluing. You obtain an open 3-dimensional manifold $N$ which is homotopy-equivalent to $S$ but not homeomorphic to an open interval bundle over $S$. The reason is that $N$ is not "tame". Proving non-tameness of $N$ requires some work to prove (by appealing to its "fundamental group at infinity"); the proof is basically the same as the one showing that $W$ is not tame.
Best Answer
As pointed out, every finitely presented group arises as the fundamental group of a closed smooth orientable four-manifold.
The same is not true of non-orientable manifolds as $\mathbb{Z}/3\mathbb{Z}$ illustrates. A necessary condition is that the group must have an index two subgroup (i.e. the fundamental group of the orientable double cover). This turns out to be sufficient. That is, a finitely presented group is the fundamental group of a closed smooth non-orientable four-manifold if and only if it has an index two subgroup; see this question.