If a ray of light at angle $\alpha$ above the horizontal hits your curve $y = f(x)$ from below at a point where the tangent to the curve has angle $\beta$ below the horizontal, it will reflect at angle $\alpha + 2 \beta$ below the horizontal, and then come back up at $\alpha + 2 \beta$ above the horizontal.
In particular, if $\alpha + 2 \beta = \pi/2$ it goes vertically down (and then retraces itself backwards), and if $\alpha + 2 \beta > \pi/2$ it goes backwards (i.e. to the left).

Let the $n$'th reflection on the curve take place at $(x_n, y_n)$, with incoming ray at angle $\alpha_n$. Then we have
$$\eqalign{\alpha_{n+1} &= \alpha_n - 2 \arctan(f'(x_n))\cr
y_{n+1} + y_n &= \tan(\alpha_{n+1}) (x_{n+1} - x_n)\cr
y_{n+1} &= f(x_{n+1})}$$
Thus $$\dfrac{\Delta \alpha_n}{\Delta x_n} = \dfrac{\alpha_{n+1}-\alpha_n}{x_{n+1} - x_n} = \tan(\alpha_{n+1}) \dfrac{- 2 \arctan(f'(x_n))}{
f(x_{n+1}) + f(x_n)} $$
In order for $x_n \to \infty$ with $\alpha_n$ increasing but staying below $\pi/2$, we would certainly need this to go to $0$. In the case $f(x) = e^{-x}$, that certainly won't happen, as $\arctan(f'(x_n)) \approx f'(x_n) = - f(x_n)$, while
$f(x_{n+1}) + f(x_n) < 2 f(x_n)$. More likely candidates would be functions
$f$ that go to $0$ very slowly, perhaps something like $1/\log(x)$.