So I thought I'd synthesize the answers everyone wrote with my understanding, and provide a breakdown of how it works depending on the angle.
Again, a reminder that this is imagining a perfectly uniform and spherical Earth. In reality, Earth's gravity field deviates from a simple $1/r^2$ profile, mostly due to oblateness but also because it's a bit "lumpy" and irregular. So there will be deviations from these simple Keplerian approximations.
So: thanks for everyone who pointed out my primary error: yes, the old and new orbits will intersect, but they won't necessarily have the same period. So the baseball will arrive back at the same position the space station was at, but probably at a different time than the station does!
It mainly depends on the direction you throw it:
Perpendicular to orbit, parallel to ground (cross-track)
Let's say you throw it off to the side. That is, you're facing "forward" (prograde), in the direction of travel, with your feet down toward Earth (nadir), and you toss it to your left or right.
Then, as PearsonArtPhoto said, it just changes the inclination of the baseball's orbit, not the period. So in half an orbit (45 minutes), it will come back to the station and hit it.
Forward or backward (prograde or retrograde)
If you throw it directly forward (prograde) or backward (retrograde), you are changing the speed but not direction of motion. This results in a new orbit, tangent to the current orbit, just like in a Hohmann transfer orbit
Let's assume for simplicity that the initial orbit is perfectly circular.
If you throw it forward, increasing the speed, the baseball's apogee will be higher, and occur 180° from the throw point, but its perigee will be at the same place you threw it from. But since the orbit is longer, it will take more time, putting the baseball back at the same point later. This means it'll appear to approach the station, but reach its closest point behind it, before moving away again.
The opposite is true for the case where it is thrown backward. The speed decreases, giving its orbit a perigee 180° from the throw point. The period will therefore be shorter and it it will appear to approach in front of the station and then recede.
In either case, eventually, after many orbits, the closest approach might come back around from the front, and it might hit the station. But it would be pretty unlikely to arrive at just the right time.
Up or down (radially)
This is the one I'm least sure of, but I think I figured out how it works. Please correct me if I'm wrong.
If you throw it "upward" (toward space, away from earth), its new apogee will be 90° ahead of where you threw it, and higher than the old one. But you didn't give it any added forward velocity, so its new, higher apogee will be "unearned", in a way. It pays for it later, 270° after the throw, where its new perigee is lower than its old one.
And what happens when it arrives back at the point you threw it? Just like throwing it prograde, the new orbit has a longer period, so it will get close to the station, but behind it.
Throwing it downward works similarly, but in reverse. It reaches a new perigee 90° after the throw, lower than the old orbit, but it's still going "too fast" for that low altitude, so at 270° it ends up at its new apogee, higher than the old one. I'm not 100% on what happens when it gets back to the point of the throw, but I assume, because of symmetry, the new orbit will have a shorter period and it'll end up ahead of the station.
I hope this is all correct; it's mostly informed by people's responses here, experience in KSP, and intuition. I made this answer a community wiki so people can make it more correct over time. I just thought it'd be good to have a place to accumulate answers for each direction.