Is there an elementary way to see that there is only one complex manifold structure on $R^2$?

There are two, not one. The open unit disc is homeomorphic to $\Bbb R^2$, and it's not biholomorphic to $\Bbb C$, so it's a distinct complex structure.

You will have trouble proving this in an easy way. Let me restate the uniformization theorem:

Every simply connected complex surface is biholomorphic to the plane, sphere, or open unit disc.

But I can tell you all of the simply connected surfaces without boundary: there's $S^2$ and $\Bbb R^2$ and that's it! So we can rephrase the uniformization theorem as follows:

Every two complex structures on $S^2$ are biholomorphic. Every complex structure on $\Bbb R^2$ is biholomorphic to either the complex plane or open unit disc.

So the uniformization theorem does not say much more than what you're asking for; if someone asked me for a proof I would tell them to read the uniformization theorem. (You can't just use Riemann mapping - it's not obvious that a complex manifold homeomorphic to $\Bbb R^2$ has a holomorphic embedding into $\Bbb C$.)