For a retrograde satellite, you are right that the satellite will migrate inwards towards the planet. Contrary to a prograde orbit, the rotation of the primary will slow down.
Think about it in terms of angular momentum. Let the primary have a positive angular momentum and the satellite a negative one (since they rotate/orbit in opposite directions). Since the satellite is being pulled inwards, its angular momentum is being lowered in magnitude (becoming less negative). In that case the primary's spin angular momentum needs to become less positive to conserve the total angular momentum. This means the primary's rotation will slow down.
An example with numbers (unitless):
Primary(initial) = 10 // Satellite(initial) = -5 ///
Primary(final) = 7 // Satellite(final) = -2 ///
Thus a transfer of 3 angular momentum units has occurred.
Eventually the satellite will slow down the primary until the primary is no longer rotating (assuming the satellite is not lost by this point). Then it will cause the primary to start rotating in the same direction as the satellite is orbiting.
Hope this helps!
EDIT:
In this paper, see equation (7), the torque on the primary due to the satellite. If we only care about the sign, we can note that $N_m \textit{~}\, (n_m -\Omega_p)$ where $n_m$ is the orbital frequency of the satellite and $\Omega_p$ is the rotational frequency of the primary. Let's take our previous example of the $\Omega_p$ positive and $n_m$ negative. This would make the torque $N_m$ negative, no matter what the magnitudes of the frequencies are. Consequently, the positive $\Omega_p$ would be reduced, signifying the slowing of the primary's rotation.
$$N_m = 3 k_2 \tau (n_m - \Omega_p) \frac{GM_m^2 R_p^2}{a_m^6} \tag{7}$$
where the subscripts $m$ and $p$ refer to the moon and planet respectively, $k2$ is the Love number of the planet that depends on its rigidity and $\tau$ is the tidal time lag of the planet.
