Classical proof of the gyromagnetic ratio $g=2$
I went through unanswered questions, and stumbled over this...
Did you find the original books?
The mistake should be in your formula for the $\mu$ of a hollow sphere; the value with $1/5$ you gave is that of a solid sphere...
The problem gets more simple I think, if you compare the two things directly:
You get both, the angular momentum and $\mu$, from highly analogous integrals over all points, in which there is a $r^2\mathrm dm$ or an $r^2\mathrm dq$:
$$ S = I\omega = \omega \int r^2\,\mathrm dm $$ and with the definition of $\mathrm d\mu$ as current times enclosed area: $$ \mu = \int\mathrm d\mu = \int A\,\mathrm dI = \int \pi r^2\cdot\frac{\mathrm dq}T = \int \pi r^2\cdot\frac{\mathrm dq}{2\pi}\omega = \frac \omega 2 \int r^2\,\mathrm dq $$
The g-factor is defined to be one if the charges coinside with the masses (the ratio of their densities is equal everywhere), i.e. the definition accounts for the $1/2$ in the second formula.
Thus, if you distribute the charge further from the axis that the mass, you get a g-factor greater than one. The integrals are always equivalent and depend on the geometry of the distribution.
For the same geometry you will always get a pre-factor for the intertia which is twice the factor for the magnetic moment -- and thus by definition a $g=1$.
Now comes the strange thing: the pre-factor in the moment of inertia of a full sphere is $1/5$ and for a hollow sphere $1/3$. The g-factor with the distribution of the mass in the sphere and of the charge on the shell thus gives a $g=5/3$.
This is obviously in contrast with the claim, that it equals two. It explains, that it is greater than one, though.
Maybe back then they could not measure $g$ so well and saw only, that it is considerably greater than one, and so could explain at least this ... ?
So the point seems to be, that the charges are further from the axis than the masses. The sphere is just a nice example, which explains the (measured) factor to be greater than one by a beautiful/plausible distribution.
...The argument with the relativistic velocities (from the comments) goes in another direction: since other measurments suggest a maximal radius for the electron, you can compute the neccessary velocities, which disproves the naive explanation of the spin (for both, the inertia and the magnetic aspect; this has nothing to do with their ratio) as a real motion.
It seems that some people liked this question so I shall post my thoughts so far. I don't have a definitive answer, but I did get some interesting results.
Let $\rho_m(\boldsymbol r)$ and $\rho_e(\boldsymbol r)$ be the mass and charge densities of the electron. The $g$ factor is given by $$ g=\frac{m}{e}\frac{\int\mathrm d\boldsymbol r\ r^2\sin\theta\ \rho_m(\boldsymbol r)}{\int\mathrm d\boldsymbol r\ r^2\sin\theta\ \rho_e(\boldsymbol r)} \tag{1} $$
From this, it's easy to see that if $\rho_m\propto \rho_e$, we get $g=1$. This means that if we have a solid sphere with constant charge density and constant mass density, the $g$ factor is 1; al hollow sphere with surface charge has also $g=1$. If we want $g\neq 1$ we must take a charge density that is not proportional to the mass density.
The first model that comes to mind is to take a volume mass density and a surface charge density, that is, a filled sphere with its charge on the surface: \begin{align} \rho_m&=\frac{m}{V}\Theta(R-r)\\ \rho_e&=\frac{e}{S}\delta(r-R)\tag{2} \end{align} where $V=\frac{4}{3}\pi R^3$ and $S=4\pi R^2$. If we plug these functions into $(1)$ we get $g=5/3$ as already acknowledged by Ilja and Anubhav. This means that Arabatzis' and Pais' claims are inaccurate: this model does not predict $g=2$ but $g=1.67$ instead.
To go a step further, we may take the same model before, but with a different mass and charge radii, that is, \begin{align} \rho_m&=\frac{m}{V}\Theta(R_m-r)\\ \rho_e&=\frac{e}{S}\delta(R_e-r)\tag{3} \end{align} with $R_m\neq R_e$. In this case, we find $g=5R_e^2/3R_m^2$, which equals 2 if $R_e=1.095 R_m$. This model seems highly artificial though.
The next possible example could be to take exponential densities, which could be the result of some kind of screening at some fundamental level: \begin{align} \rho_m&\propto\exp\left[-\frac{r^2}{R_m^2}\right]\\ \rho_e&\propto\exp\left[-\frac{r}{R_e}\right]\tag{4} \end{align} from which we find $g=8R_e^2/R_m^2$; if we take $R_m=2R_e$ we get $g=2$. This is still very artificial but there might be some electrostatic model that is able to accommodate this.
Other possible models could consist of non-spherical densities, such as cylinders or string-like wires. I leave to the reader to explore this models. In any case, it is clear that the most natural models don't predict $g=2$, and it's not easy to find another one that fixes this while not getting too ad-hoc. But it is possible to write down exotic models with tunable parameters so as to get $g=2$, which means at least that $g=2$ is achievable at the classical level.