Now that's a mildly non-trivial observation. Why would they be equal, really?
Let's say a particle with mass $m$, charge $q$, and initial velocity $v$ enters an area of length $L$ where an electric field $E$ starts to deflect it sideways. This is a clear example of uniformly accelerated motion, and its laws are well known: $x=vt,\;y={at^2\over2}$, where the acceleration $a={qE\over m}$. By the time the particle reaches the right side of the picture, it gets deflected by $y={at^2\over2}={1\over2}{qE\over m}\left({L\over v}\right)^2={1\over2}{qEL^2\over mv^2}$.
That pretty much sums it up. $E$ and $L$ are the same for everyone, $q$ is the same in absolute value for electron and proton, so the only difference is $mv^2$. Now, considering your two situations:
- Same speed: same $v$, but a proton is slightly (that is, a thousand-and-something times) heavier, hence lower deflection.
- Same energy: same $mv^2$, same deflection (just to the opposite sides).
Things get hairy in relativity, but that's another story.
So it goes.