Consider an elliptic curve $y^2=x^3+ax+b$. It is well known that we can (in the generic case) create an addition on this curve turning it into an abelian group: The group law is characterized by the neutral element being the point at infinity and the fact that $w_1+w_2+w_3=0$ if and only if the three points $w_j$ are the intersections (with multiplicity) of a line and the elliptic curve.

Groups can be hard to work with, but in most cases proving that the group is in fact a group is easy. The elliptic curve is an obvious exception. Commutativity is easy, but associativity is hard, at least to this non-algebraist: The proof looks like a big calculation, and the associativity seems like an algebraic accident rather than something that ought to be true.

So this is my question, then: Why is the group law on an elliptic curve associative? Is there some good reason for it? Is the group perhaps a subgroup or a quotient of some other group that is easier to understand? Or can it be constructed from other groups in some fashion?

I gather that historically, the group law was discovered via the addition law for the Weierstrass $\wp$-function. The addition law is itself not totally obviuos, plus this approach seems limited to the case where the base field is $\mathbb{C}$. In any case, I'll elaborate a bit on this shortly, in a (community wiki) answer.

Everything I am writing below is carried out explicitly in Chapter III of Silverman's book on elliptic curves. In the earlier chapters, he defines the Picard group.
For any curve over any field, algebraic geometers are interested in an associated group called the Picard group. It is a certain quotient of the free abelian group on points of the curve. It consists of formal sums of points on the curve modulo those formal sums that come from looking at the zeroes and poles of rational functions. It is a very important tool in the study of algebraic curves.
The very special thing about elliptic curves, as opposed to other curves, is that they turn out to be in natural set-theoretic bijection with their own Picard groups (or actually, the subgroup $\operatorname{Pic}^0(E)$). The bijection is as follows: let $O$ be the point at infinity. Then send a point $P$ on the elliptic curve to the formal sum of points $[P] - [O].$ (It is not obvious that this is a bijection, but the work to prove it is all "pure geometric reasoning" with no computations.) So there is automatically a group law on the points of $E.$ Then it requires no messy formulas to show that under this group law, the sum of three collinear points is $O.$ So for free, you also get that this group law is the same as the one you defined in the question and that the one you defined is associative!
