From the Classification Theorem for closed (i.e. compact and boundaryless) surfaces, it follows that $S^2$ is the only closed surface with trivial $\pi _1$. That's easy because the fundamental group classifies closed surfaces.
I'd like to get the same conclusion (i.e. that a simply-connected closed surface is isomorphic to $S^2$, in your favorite category) without that theorem. Is it possible? Any discussion about (generalized) Poincaré conjecture I could find starts saying that in dimension $1$ and $2$ it's true because of the classification theorems (available in those dimensions), dimension $3$ was settled by Perelman and then switches to high-dimensional wild cases. No insights on a direct proof in dimension $2$.
My attempt: let $S$ be a closed surface with $\pi_1(S)=1$; trivial $\pi_1$ implies that $S$ is orientable, that $S$ is covered only by itself, that every embedded loop bounds a disk (so $S$ has genus $0$, since cutting along every embedded loop disconnects $S$); compactness rules out $\mathbb{R}^2$ and $\partial S = \emptyset$ rules out the unit disk. But here I don't know how to go on. Maybe one should try to find an explicit isomorphism to $S^2$, but I can't see how.
EDIT: Of course (as suspected), this argument is wrong. The problem is that I may well have lots more critical points, as long as they are canceled out by critical points of index 1. Some massaging is necessary, but then the proof is less fun. Anyway, I'll leave this here in case anyone can find a Morse theoretic approach. (And make it community wiki so they can add it.)
This might go against the spirit of the question, but it's fun anyway. (Assuming it's correct; I have a tendency to be careless at this hour...)
If you're willing to assume a little Morse theory and that the surfaces are smooth than this is immediate:
Pick a Morse function. The indices of the critical points can be either 0, 1, or 2. But you know by the Morse inequalities that there is at least one critical value of index 0 and one of index 2 (since you know the Betti numbers). Moreover, the Euler characteristic of the surface is 2, so by the Morse inequalities again you actually know that there are precisely two critical points- one of index 0 and another of index 2.
Morse theory (Reeb's theorem) gives us a construction showing that manifolds with Morse functions having exactly two critical points must be homeomorphic to a sphere.
It's plausible that with enough unwraveling the above argument could be made elementary... after all- everything is in 1 and 2 dimensions so one can draw pictures! :)