Added. I haven't quite found the old reference yet, but here is the proof. If R is a ring whose multiplicative reduct is completely regular, then it is von Neumann regularb and so its Jacobson radical is trivial. Thus it suffices to handle the primitive case, so assume R has a faithful simple module. Clearly 1 and 0 are then the only central idempotents of R. Now we show all idempotents of R are central. By Clifford's structure theorem for completely regular semigroups, R is a semilattice of completely simple semigroups. So it suffices to show these completely simple semigroups are groups and then R will be an inverse semigroup with central idempotents. For this it suffices to show $\mathcal R$-equivalent and $\mathcal L$-equivalent idempotents e,f are equal. By symmetry we assume eR=fR. Then $(e-f)^2=0$ and hence e-f=0 since R is completely regular.
Thus R is a completely regular inverse monoid whose only idempotents are 0,1 and so R-{0} is a group, i.e., R is a division ring.
I believe the old paper I can't find shows a ring R is completely regular iff it is what ring theorists call strongly regular.
