Edit: I should have put a short version of the answer in the beginning, so here is how the various structures are recovered. To get a smooth manifold structure on the quotient, you use the fact that $SL_2(\mathbb{R})$ is a real Lie group and $SO_2(\mathbb{R})$ is a closed subgroup. To get a hyperbolic structure, you use the fact that $SL_2(\mathbb{R})$ is isomorphic to an orthogonal group of signature (n,1) for some n (giving a transitive action on hyperbolic n-space). To get a complex structure, you use the fact that $SL_2(\mathbb{R})$ is isomorphic to an orthogonal group of signature (2,m) for some m (giving an action on a hermitian symmetric space).
As others have noted, you can get a bijection on points using the Iwasawa decomposition, and you can get a hyperbolic structure using the exceptional isomorphism $PSL_2(\mathbb{R}) \cong SO_{2,1}^+(\mathbb{R})$. First, I'd like to clean up the Iwasawa treatment a bit. Any element of $SL_2(\mathbb{R})$ can be uniquely decomposed as BK, where K is a rotation and B is upper triangular with positive diagonal. Any rotation K fixes i, so we should consider what elements B do. A bit of fiddling shows that $\begin{pmatrix} \sqrt{y} & x/\sqrt{y} \\ 0 & 1/\sqrt{y} \end{pmatrix} \cdot i = x+iy$.
We can view the exceptional isomorphism in another way that makes the complex structure more apparent, by viewing the hyperbolic plane as the Grassmannian $O_{2,1}(\mathbb{R})/(O_2(\mathbb{R}) \times O_1(\mathbb{R}))$. From the standpoint of special relativity, this is the space of timelike lines through the origin in $\mathbb{R}^{2,1}$. Taking a quotient of the total space of these lines (minus origin) by positive rescaling, we find that this space is isomorphic to the space of pairs of antipodal points of norm -1. In particular, we have an isomorphism of the Grassmannian with the quotient of the hyperboloid with two sheets (i.e., solutions of the equation $x^2 + y^2 - z^2 = -1$) by the antipodal automorphism.
One way to explain the origin of the complex structure is by the fact that all Grassmannians of the form $O(2,n)/(O(2) \times O(n))$ are hermitian symmetric spaces, and the hyperbolic plane is just the case $n=1$. The 2 in $O(2)$ is essential, because the orthogonal group action is what yields the ninety degree rotation in the tangent space of any point, and this is what endows the quotient with an almost complex structure. If you want to see more about hermitian symmetric spaces than the Wikipedia blurb, I recommend looking in chapter 1 of Milne's introduction to Shimura varieties.
Finally, I'd like to point out Deligne's description of the upper half plane as a moduli space of structured elliptic curves. Points on H parametrize elliptic curves with an oriented basis of first homology (as mentioned a few times in our class). If you want to say it is a fine moduli space, you need a functor that it represents, and it is unfortunately a bit complicated. The functor takes as input the category of complex analytic spaces, and for any such space S, it gives the set of isomorphism classes of elliptic curves over S (i.e., diagrams $E \underset{\pi}{\leftrightarrows} S$ of complex analytic spaces, where $\pi$ is smooth and proper with one-dimensional genus one fibers and the leftward map is a section) equipped with an isomorphism $R^1\pi_*\underline{\mathbb{Z}} \cong \underline{\mathbb{Z} \times \mathbb{Z}}$ that induces the canonical identity $R^2\pi_*\underline{\mathbb{Z}} \cong \underline{\mathbb{Z}}$ on exterior squares. Here, the underscore indicates a constant sheaf. The functor also takes morphisms to "the evident diagrams". To be honest, I have never seen a complete proof that this functor is represented by the complex upper half plane, although it seems to be more a question of doing lots of writing than an honest theoretical problem. You can probably do it using the fact that H is a classifying space of polarized Hodge structures, as Kevin Buzzard mentioned in the comments.
