Geometric notion of addition for the real projective line

Draw a vertical line $\ell$ that meets both lines $a$ and $b$. (This is always possible: if one or both of the lines is vertical, use it.) Pick a point $p$ common to $a$ and $\ell$, and similarly a point $q$ common to $b$ and $\ell$. ($p$ is unique unless $a$ is vertical, in which case it is any point on $a$; the same goes for $q$ and $b$.) Construct a third point $r$ on $\ell$ whose $y$-coordinate is the sum of those of $p$ and $q$. The line passing through the origin and $r$ is $a+b$.

I've just found another construction that satisfies the given requirements. I know it comes one year late, but I'm posting it here as a supplementary solution just in case someone finds it interesting. The construction is based on the one for multiplication that I outlined in the question.

We work in $\mathbb{R}^4$, coordinatized by $(x,y,z,t)$. The starting configuration is a line $a$ in the plane XY and a line $b$ in the plane XZ (notice the slight difference with the case of multiplication; the shared axis here is the X axis).

To construct $a+b$, one extends the line $a$ perpendicularly to a hyperplane in $\mathbb{R}^4$, then does the same with the line $b$, then finds the triple intersection between these two hyperplanes and a third hyperplane given by the equation $y+z-t=0$. Finally, one projects this intersection onto the plane XT to find the desired sum.

It is straightforward to check, for example, that the lines with finite slope given by the equations $$a: \begin{cases} y=m_1 x\\ z=0\\ t=0\end{cases}$$ in the plane XY and $$b: \begin{cases} z=m_2 x\\ y=0\\ t=0\end{cases}$$ in the plane XZ correctly sum to the line $$a+b: \begin{cases} t=(m_1+m_2) x\\ y=0\\ z=0\end{cases}$$ in the plane XT. Similarly, the construction gives $\infty + a = a + \infty = \infty$ for any $a$ as required (including $\infty$).

This construction has the disadvantage that it can't be easily visualized like the others, since it requires four dimensions. However, it has the advantage of not requiring any arbitrary choices, and it still works if one replaces the lines with the graphs of arbitrary functions in $\mathbb{R}^2$. In that sense, it nicely supplements the accepted answer from Rahul.