Building off of Rob Jeffries answer, I will try to write out a complete expression for the case of electrons with higher kinetic energy than the Blackbody thermal energy $\sim k_b T$.
The starting point is the expression for the total power irradiated by an electron to a photon density $U_{\mathrm{photon}}$ through inverse Compton scattering (see the nice reference https://casper.ssl.berkeley.edu/astrobaki/index.php/Inverse_Compton_Scattering).
$$P_{\mathrm{tot}}=\frac{4}{3} \sigma_T U_{\mathrm{photon}} c \beta^2 \gamma^2$$
Where $\sigma_T$ is the Thomson cross-section $\sigma_T = \frac{8\pi}{3}(\frac{\alpha \hbar c}{m c^2})^2 \approx 66.5 \,\mathrm{fm}^2$, $U$ is the photon density, which for blackbody radiation is simply $U=\frac{4 \sigma T^4}{c}$, where $\sigma$ is the Stefan-Boltzmann constant.
We can write $P_{\mathrm{tot}}$ as a function of (kinetic) energy by noting $KE=(\gamma-1)m c^2$ and $\beta^2 = 1-1/\gamma^2$.
$$P_{\mathrm{tot}}=\frac{4}{3} \sigma_T U c \frac{KE\,(KE+2 mc^2)}{m c^2}$$
Then because $P_{\mathrm{tot}} dt = dKE$, the total time taken to radiate from the starting (kinetic) energy $E_0$ to the thermal energy $\frac{3}{2} k_b T$ is given by the integral:
$$\tau = \int_{E_0}^{\frac{3}{2} k_b T} \frac{1}{P_{\mathrm{tot}}} dKE$$
The end result of this is:
$$\tau = \frac{m c^2}{\frac{8}{3} \sigma_T U c} \mathrm{log}\left[ \frac{E_0+2m c^2}{\frac{3}{2} k_B T + 2m c^2} \frac{\frac{3}{2} k_B T}{E_0}\right]$$
Plugging in all the constants and using units of Kelvin for the temperature $T$, we get
$$\tau \approx \frac{1}{T^4} \mathrm{log}\left[ \frac{E_0+2m c^2}{\frac{3}{2} k_B T + 2m c^2} \frac{\frac{3}{2} k_B T}{E_0}\right] \times2\times 10^{21} \,\,\mathrm{Seconds}$$
The weak logarithmic behavior means for practical purposes the logarithm only changes by a factor of about 3 for all electron kinetic energies above $k_b T$.
So this means it would take $\sim 10^{21}$ seconds for any electron to equilibriate with the CMB blackbody radiation, which is much longer than the age of the universe. On the other hand, if the electron was traveling through an empty region at 10,000 Kelvin, it would equilibriate in $\sim 10^{6}$ seconds, which may be of some astrophysical relevance.