There is another solution using algebraic geometry. If you consider that set of ordered pairs (point on the ellipse, line passing through the point on the ellipse and at least one focus), you see that this set defines an algebraic curve $C$. The natural map from $C$ to the ellipse (that we get by forgetting the line) is two to one at every point except for the two points where the line through the foci hits the ellipse, so by the Riemann-Hurwitz formula $C$ has genus zero, i.e. it is isomorphic to the projective line.
You have defined a map $f$ from $C$ to itself, and this map can be defined algebraically and is clearly invertible, so as a map from the projective line to itself it is a fractional linear transformation. The map $f \circ f$ has two obvious fixed points where the line through the foci intersects the ellipse, and if we fix an identification of $C$ with the projective line sending the two fixed points to $0$ and $\infty$, then we see that the map $f$ must have the form $f(x) = c/x$ for some constant $c$ (depending on the eccentricity of the ellipse). Thus as we iterate $f$, $f^n(x)$ tends to $0$ and $\infty$ (in other words, the billiard ball tends to a horizontal line), and it does so at an exponential rate.