ac.commutative algebra - Why is an elliptic curve a group?

Short answer: because it's a complex torus. Explanation below would take as through many topics.

Topological covers

The curve should be considered over complex numbers, where it can be seen as a Riemann surface, therefore a two-dimensional oriented closed variety. How to find out whether this particular one is a sphere, torus or something else? Just consider a two-fold covering onto $x$-axis and count the Euler characteristics as $-2 cdot 2 + 4 = 0$ (don't forget the point at infinity.)

Complex tori

So this is a torus; now a torus with complex structure can be always defined as a quotient $mathbb C/Lambda$, where $Lambda$ is the lattice of periods. It can be written as integrals $int_gamma omega$ of any differential form $omega$ over all elements $gamma in pi_1$. The choice of differential form is unique up to $lambda in mathbb C$.

Algebraic addition

A complex map of a torus into itself that leaves lattice $Lambda$ fixed can be only given by a shift. Once you select a base point, these shifts are in one-to-one correspondence with points of $E$. We have unique distinguished point — infinity — so let's choose it as the base point. It follows that we now have an addition map $(u, v) to uoplus v$, though defined purely algebraically so far.

Geometric meaning

Now let's stop and ask ourselves: how to see this addition geometrically? For a start, consider map that sends $u$ to the third point of intersection with the line containing both $u$ and 0 (the infinity point). It's not hard to see that we fix 0 but change every class $gamma$ in a fundamental group into $-gamma$, so we must have the map $umapsto -u$ here.

Group theory laws

What would happen if you took a line through $u$ and $v$? By temporarily changing coordinates so that $u$ becomes the infinity point, one writes down that map as $(u, v) mapsto -(u+v)$.
Now if you took three points, there would be two different ways to add them; those would lead to $(u+v)+w$ and $u+(v+w)$ as complex numbers, which we know to be associative.

Logically proven

In the above, we worked over complex numbers, but we proved associativity which is a formal theorem about substitution of some rational expressions into others. Since it works over complex fields, it is required to work over all fields.

(In any case, the big discovery of mid-20th century was that you actually can take all of the intuition described above and apply it to the case of elliptic curves over arbitrary field)

Analytic computations (bonus)

Consider a line that passes through points $u$, $0$ and $-u$. This line is actually vertical, and $y$ is a well-defined function there which has two zeroes and one double pole at infinity. After a shift and multiplication of several such functions we'll be getting a meromorphic function on a complex torus with poles $p_i$ and zeroes $z_i$ having the property $sum p_i = sum z_i$. This method can give all such functions and only them; it's not hard to see that only meromorphic functions with this property are allowed on elliptic curve.

For example, $wp'$-functions are the ones that have triple pole at 0 and single zeroes at points $frac12w_1, frac12w_2, frac12(w_1+ w_2)$ where $w_1, w_2$ are generators of $Lambda$.

Jacobian of a curve (bonus 2)

The formula above describes what types of functions are allowed on our curve. It is a good idea to organize this information into a curve: in this case, the information is that a single expression $p_1 + p_2 + cdots + p_n - z_1 - cdots - z_n$, considered a point of the curve, must vanish. For curves of higher genus, more relations are necessary; for $mathbb Cmathbb P^1$, no relations beyond number of poles = number of zeroes are necessary. Those are relations in the group of classes of divisors (= Jacobian of a curve) mentioned in other answers.

In particular, elliptic curves coincide with their Jacobian and that's another explanation for the additive law.