There is another effect that can result in additional spin drag in the form of viscous drag.

Preamble: both bullets fly with the exact same speed and perfect orientation, so zero angle of attack.

As you elaborated, the viscous friction in the spinning case is higher. This can be additionally contributed by an earlier boundary transition from laminar to turbulent.

In the spinning case, the flow spirals around the bullet, effectively traveling a longer distance in the same amount of time. The relative velocity between surface and air is higher. This increases the Reynolds number, (rho * v * l_ref / eta; rho, l_ref & eta are constant between the cases) possibly leading to earlier laminar-turbulent transition. Also, the flow distance along the spiral is bigger, so there's more distance for the transition to happen. If it were to happen at the same run length of the boundary layer, due to the spiral the transition would effectively happen earlier along the bullet spin axis. And since turbulent boundary layers exert more drag than laminar ones, this can increase drag for the spinning bullet.