Ignore that; I’m an idiot. If you treat it abstractly, it’s almost blindingly obvious.
You just have to prove that L^2 / A is constant within a similarity class. Take two members of the same similarity class, of areas A and A’ and lengths L and L’. Let F be the factor of the dilation that maps the first figure onto the other. Then A = F^2 * A’ and L = F * L. Squaring the length equation gives L^2 = F^2 * L’^2. Dividing the area equation by this, the F^2 factors cancel, yielding A / L^2 = A’ / L’^2. So the area to squared length ratio is indeed constant.