Consider the integral:
$ h(r, \theta) = \frac{1}{2\pi} \int_0^{2\pi} g(\phi) \frac{1-r^2}{1-2r \cos(\theta - \phi) + r^2} d\phi, r<1$
I want to show that $\Delta h=0$, but in order to do so, I need to justify this:
$\Delta \frac{1}{2\pi} \int_0^{2\pi} g(\phi) \frac{1-r^2}{1-2r \cos(\theta - \phi) + r^2} d\phi = \frac{1}{2\pi} \int_0^{2\pi} g(\phi) \Delta \frac{1-r^2}{1-2r \cos(\theta - \phi) + r^2} d\phi$
That way, this becomes $0$, as the Poisson kernel is harmonic.
I think I solved the problem. Seems like I can somehow relate the function to a regular analytic complex function, as shown in the last few pages of this: https://www.math.uh.edu/~shanyuji/Complex/Notes/N27.pdf