Units for a PSD from square of FT?
Your reasoning sounds correct (with unit considerations), you can see a discussion on both continuous and discrete aspects of the Fourier transform in What are the units of my data after an FFT? 

What is unclear to me is why  you expect the PSD to be in unit $\frac{m^2}{Hz}$? Powers are powers: if a quantity has unit $\textrm{u}$, its square has unit $\textrm{u}^2$. However, when dealing with physics, the quantities you mention are often somehow periodic, and evaluated via averaging over one period, or asymptotically. So basically you had a normalization factor, proportional to the inverse of a time duration, or $\textrm{s}^{-1}$. Hence, possibly, your $\textrm{Hz}$ factor.