How does the Fourier Transform invert units?

If you integrate $$\int f(x)\textrm{d}x = F$$ then $F$ has the units of $f$ times the units of $x$.

Similarly if you differentiate, $$\frac{\textrm{d}f}{\textrm{d}x}$$ has units of $f$ divided by units of $x$.

If you look at the simple example of integrating and differentiating with respect to time to go between position, velocity, and acceleration, you'll see this works. For example, differentiate position (meters) with respect to time (seconds) to get velocity $(\frac{\textrm{m}}{\textrm{s}})$.

If you have a function of time and you Fourier-transform it, and then perform the inverse $$f(t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} F(\omega)e^{-i\omega t}\textrm{d}\omega$$ you should have the same units on both sides. $\omega$ has units of $\textrm{s}^{-1}$ (and so $\textrm{d}\omega$ has the same) and $e^{-i\omega t}$ is dimensionless. That means $F$ needs to have the same units as $f$, but multiplied by $\textrm{s}$. So the Fourier transform of a time series does not have units of frequency. It actually has units of inverse-frequency added on, so that when we multiply it by a frequency, we get the same units back that we started with. For example, if $f$ is measuring volts, $F$ has units of volts-seconds. Note that we are not transforming the units of $t$ and $\omega$ at all. We are choosing to express a function either in terms of time or in terms of frequency, then figuring out what units follow along.

Yes, Fourier transforms can be applied to data aside from time-series. For example, we can Fourier-transform a spatial pattern to express it in wavenumber-space, that is, we can express any function of space as a sum of plane waves. Physically, this Fourier transform is performed (for example) by a diffraction grating, which Fourier-transforms the spatial pattern of the grating.