There's probably other ways of doing this, but I've found this to be the simplest way (group law) that does indeed work:
To add points $A, B \in \{(x, f(x)) : x \in \Bbb{C}\} = G$ where $f$ is any parabola with vertex $E \in G$, we treat $E$ as zero. Now draw a line between $A, B$ and then draw a parallel line to this line that passes through $E$. The unique intersection point (other than $E$, unless of course $A = B = E$ or $A = -B$) is then the value of the group law.
I've checked all the axioms of a group using Geogebra. This also works on a circle if I recall correctly.
I'm wondering:
How do we express $AB$ the abelian group law algebraically?
Nice observation! You can just crank it out algebraically. Starting with two points $(a, a^2), (b, b^2)$ on the parabola $y = x^2$ (to keep things simple; we can reduce to this case WLOG by a suitable change of variables), the line between them has slope $\frac{a^2 - b^2}{a - b} = a + b$, and the line with this slope passing through the origin intersects the parabola at the origin and at $(a + b, (a + b)^2)$.
So this does give us an algebraic group although it is just the familiar additive group $\mathbb{G}_a$. The interesting thing is that this construction actually makes sense on any nondegenerate conic over any field. For example on the circle $x^2 + y^2 = 1$ we recover $SO_2$ and on the hyperbola $xy = 1$ we recover the multiplicative group $\mathbb{G}_m$. The analogy to elliptic curves is discussed in detail in Lemmermeyer's Conics - a poor man's elliptic curves.