I just want to make sure that I have the correct definitions for things. I claim that $\mathbb P^1(\mathbb R)$ is an abelian variety, which according to Wikipedia is a projective algebraic variety which is also a group, where the group law is a regular map. First, is it correct that $\mathbb P^1(\mathbb R) = (\mathbb R^2 \setminus \{0\}) / \sim$, where $(a_1, a_2) \sim (b_1, b_2)$ if there exists some $c \in \mathbb R \setminus \{0\}$ such that $(a_1, a_2) = c(b_1, b_2)$?
Anyway, $\mathbb P^1(\mathbb R)$ is obviously a projective variety, given by $V(0) \subseteq \mathbb P^1(\mathbb R)$, so the first condition holds.
To find a regular map to be the group law on $\mathbb P^1(\mathbb R)$, I considered the group law to be composition, where I interpret each point $[x : y]$ to be a rotation that brings $[1 : 0]$ to $[x : y]$. Thus I get the group law $$[x_1 : y_1] + [x_2 : y_2] = [x_1 x_2 - y_1 y_2 : x_1 y_2 + x_2 y_1].$$ This law is commutative and associative, has an identity $[1 : 0]$, and has inverses, since $[x : y] + [x : -y] = [1 : 0]$. I believe it is also a regular map, since the components of $[x_1 : y_1] + [x_2 : y_2]$ are given by polynomial functions of $x_1, y_2, x_2, y_2$. Thus this law makes $\mathbb P^1(\mathbb R)$ an abelian group, so it is an abelian variety.
Is all of this right? Please tell me if I'm missing anything or misunderstand any of the definitions.
In the now antiquated language of "classical varieties", everything you wrote would be correct, except that using this language requires that your maps make sense when they take values over a fixed algebraically closed base field. In this case, note that $(i,1) + (i,-1) = (0,0)$, and $(0,0)$ is in neither $\mathbb{P}^1(\mathbb{R})$ nor $\mathbb{P}^1(\mathbb{C})$.
What you've defined is a topological group structure on $\mathbb{P}^1(\mathbb{R})$, where the group operations are "polynomial functions", but the definition of an abelian variety requires more.