And to add to that, not only does Kepler's third law about the total motion (its period and semimajor axis) apply, so does Kepler's first and second laws! The laws all apply to the motion of the displacement vector between the two bodies (so now we have left high school but you said you wanted more advanced information, the displacement vector is nothing more than an arrow that extends between the two objects, it only has a length and a direction, there's no necessary meaning to where it is anchored). All you have to do is imagine that one body with a gravitational mass equal to the total mass M1 + M2 is "nailed down", and the displacement vector connects it to the other body, moving under that gravity in exactly the way an object would if the other one was not able to move. Indeed, this is precisely the situation where all three Kepler's laws apply.
Digging a little deeper, we should notice that in binary stars, we can't really assume either star is "nailed down" (the way we invariably do for the planets orbiting a "nailed down" Sun, which is the language we use with Kepler's laws in our solar system). Instead, both stars orbit the center of mass, so does this mean our new version of Kepler's law is only approximate, whereas it is accurate in the solar system?
No, all three Kepler's laws still completely apply to the displacement vector between the two bodies. What's more, you can even get total quantities like the potential energy, kinetic energy, and total angular momentum if you imagine that the orbiting body at the outer edge of the displacement vector (the "head" of the arrow, if you will) has a mass equal to the "reduced mass", M1 * M2 / (M1 + M2). Everything is exactly like you were taught about Kepler's laws, except you pretend you have an object of mass M1 * M2 / (M1 + M2) orbiting a nailed-down object of gravitational mass M = M1 + M2. You get right all the total energies and the angular momentum too. The only thing you get wrong is the total momentum, which is not zero for the orbiting reduced-mass object (but of course is zero in the center of mass frame for the real system, so it's obvious you are getting that wrong and wouldn't need to try to use that quantity anyway).
And here's a bonus, you even get one thing right about Kepler's first law that is wrong in the usual way you were taught it. I'll bet you were taught that orbits are ellipses with the Sun at one focus. That's only approximate, and works because the Sun's mass is so much more than the planets (so it's no good for binary stars). As pointed out above, the Sun is not at the focus of the ellipse of Jupiter's orbit, it is the Sun-Jupiter (ignoring all the other planets for simplicity) center of mass that is the focus of both Jupiter's ellipse and the Sun's ellipse (which of course are very close to circles, but didn't have to be). Also, neither of those ellipses have their semimajor axis equal to the maximum distance between the objects. But if you use the description where all you track is the displacement vector between the two stars in the binary (or the Sun and Jupiter), then that vector traces out a single ellipse whose semimajor axis is the maximum distance between the objects, and the "nailed down" hypothetical object really is at one focus, unlike the Sun.
In short, this is actually the situation you have been taught all along, but didn't know it until you started thinking about binary stars.
To dig even deeper, we are always allowed to "nail down" (in effect) one of the objects, simply by choosing a coordinate system where that object is at the origin of the coordinates (which is the only thing that nails it down, it's not something physical just a choice of perspective) and if you do that, these objects with the displacement vector between them are not hypothetical at all. That is not an inertial reference frame (which is why the total momentum is not conserved in that frame), but what we are saying is you can handle all the tricky non-inertial business by completely ignoring it, if you simply treat the object that is nailed down as if it had a gravitational mass of M1 + M2 (and an effectively infinite inertial mass, as it's nailed down by the coordinates), and the object that is orbiting as if it had a gravitational and inertial mass of M1 * M2 / (M1 + M2). This is a wonderful simplification of the two-body problem, and it works for any M1 and M2. As mentioned, you even get the total energies and angular momentum right, which is quite convenient.