What colour is an electron?

I think to first order what you are looking for is the Klein Nishina cross-section. What is important here is that light can inelastically scatter from electrons, but can never be absorbed by a free electron. So instead of describing the color an electron by its absorption as a function of wavelength, you're really using its Raman spectrum, albeit at very high energies (x-rays and gamma rays instead of optical light).

https://en.m.wikipedia.org/wiki/Klein%E2%80%93Nishina_formula

In any case, the key quantity to consider is the ratio of the photon energy and electron mass of 512 keV. Well below this energy you have wavelength independent Thomson scattering (i.e. the electron is white and scatters all colors), but near the electron mass you have Compton scattering which is inelastic and angle-dependent.

At much higher energies compared to the electron rest mass, you get a roughly wavelength independent response once again. However, I would guess that other contributions will dominate the behavior depending on the exact energy scale (e.g. Schwinger limit for non-linear electrodynamics in the extreme limit).

At a more philosophical level, I think you can say the electron has a color only in the same sense that thin films or nanostructures have "color". In all these cases, it isn't the absorption of light giving the color, but instead it is the wave mechanics of the incoming and outgoing light.

Since the electron scattering cross section is purely given by the direction of the observer relative to the light source, it is very similar to something like the color of a wing of a butterfly, which also is geometry dependent.

Edit: For completeness, I will just repeat the exact formula of the scattered light energy vs incoming energy and angle below.

$$E_i - E_f = \Delta E = E_i\left(1-\frac{1}{1+\frac{E_i}{m_e c^2}\left(1-\mathrm{cos}(\theta)\right)}\right)$$

The above equation shows that the "color" of the electron depends on the exact wavelength of the illumination light and on the angle of the observer with the respect to the source. Moreover, the intensity of the inelastic light (i.e. light with different color than the illumination) is quite weak, and goes to zero away from the electron mass of 512 keV where you instead get wavelength-independent scattering again.