Why is the return current through a printed circuit board's ground plane concentrated below the circuit trace?
Remember that the story we tell children about electric currents --- that energy in electric circuits is carried by moving electric charges --- is somewhere between an oversimplification and a fiction. Energy flow in electric circuits is carried by electromagnetic fields. That's true even for DC circuits, like the one below. You can think of the transmission line between the voltage source and the load as a capacitor, with an electric field $\color{red}{\vec E}$ pointing from the upper wire down to the lower, and as an inductor with a magnetic field $\color{green}{\vec H}$ pointing into the page in the loop and returning outside. The magnitude and direction of power flow is given by the Poynting vector $\color{blue}{\vec S} \propto \color{red}{\vec E}\times\color{green}{\vec H} $, which points out from the battery, down the transmission line, and into the resistor. If you do the hard, three-dimensional integral to find out how much of the power flow in this sort of circuit is described by the Poynting vector, you find ... all of it.
By Chetvorno - Own work, CC0, Link
To find the distribution of charges on the ground plane, we have to satisfy the boundary conditions for the electromagnetic field at the surfaces of the conductors. Let's do a static charge on the conducting trace (for example: a constant-voltage signal to some FET with very large input impedance, so that it draws no current) first, to warm up. For a point charge above an infinite conducting plane at $z=0$, the method of image charges gives charge density on the ground plane
$$ \sigma = -\epsilon_0 \frac{\partial V}{\partial z} \Bigg|_{z=0} = \frac{-q a}{2 \pi \left(\rho^2 + a^2\right)^{3/2} } \tag 1 $$
where $\rho$ is the cylindrical distance from the point charge. Note that the total distance between a point $(\rho,\phi,0)$ on the ground plane and the point charge at $(0,\text{anything},a)$ is $r=(\rho^2+a^2)^{1/2}$. If $\theta$ is the angle between $\vec r$ and the vertical, we have $\sigma \propto r^{-3}a \propto r^{-2}\cos\theta$ for a point charge.
We got that result using the potential for a point charge (and its mirror image) of $V\propto r^{-1}$ for a distance $r$ from the charge. The potential due to a line charge goes like $V\propto \ln r$ for a (perpendicular) distance $r$ from the charge. Doing the same derivative as in (1) ought to then give us $\sigma \propto r^{-1}\cos\theta$, but I'll leave the details and $2\pi$s and so on to you.
So there's your derivation for a static line charge. How does the charge distribution on the ground plane change if there's a current? The answer is that it doesn't. The boundary condition on $\vec E$ at the ground plane is still that $\vec E$ must be normal to the plane. The current introduces everywhere a magnetic field $\vec B$ that's perpendicular to the direction of the line current. Consider the DC/zero-frequency case first: there's no $\partial\vec B/\partial t$ to change the electric field, so the charge density must be the same as in the static case. This rebuts the (reasonable) intuition you state in your question, that the ground-plane current should be magnetically repelled by the line current: doing so would set up a transverse electric field on the surface of the ground plane conductor.
For an AC current, you can think of the current trace as a line charge whose charge density varies sinusoidally along its length. The method of image charges suggests that the surface charge under the ground plane should still vary like $\sigma\propto r^{-1}\cos\theta$ for any "transverse slice" under the current trace, giving the same electric fields as if there were an anti-trace below the ground plane with the opposite charge distribution. That charge distribution does have nonzero $\vec E_\parallel$ at the ground plane, in the direction of the line current; those are the fields that, together with the changing $\vec B$, move the ground-plane charges in the direction parallel to the current flow. The job of proving that the ground-plane component of $\vec E$ transverse to the circuit trace vanishes everywhere I'll leave to you as well.
Here's another argument that the static and dynamic charge distributions transverse to the circuit trace should be the same, based on symmetry. (I think I learned about this problem from Griffiths's E&M textbook.) Imagine you have two infinite, parallel, opposite-sign line charges at rest relative to each other: they'll experience an electrostatic attraction, but no magnetic interaction. However, another experimenter walking past you sees the two line charges, in her boosted reference frame, sees the line charges as antiparallel currents, which have a magnetic repulsion in addition to their electrostatic attraction; the magnetic repulsion becomes stronger, in her reference frame, as she passed you more quickly and the apparent current becomes larger. It seems plausible that, if the other experimenter walks past you sufficiently rapidly, she might see the magnetic repulsion become larger than the electrostatic attraction. Then you would disagree about whether the two line charges were attracted or repelled from each other. It is instructive to compute the speed of the boost at which the magnetic and electric forces balance each other exactly --- though if you've studied special relativity you might be able to guess the answer.