The question implies that when the disc is initially put down it is not rolling without slipping yet. In fact, at $t=0$ it's not yet rolling at all.
Simple conservation of momentum doesn't apply here because friction energy is not conserved.
The disc exerts a force $mg$ (its weight) on the surface, which in turn exerts an equal but opposite Normal force $F_N$, which prevents the ring from penetrating the surface:
$$F_N=mg$$
In the simple friction model a friction force $F_f$ is exerted in the opposite sense of motion, acc.:
$$F_f=\mu F_N=\mu mg,$$
where $\mu$ is the friction coefficient.
We can now set up two equations of motion:
1. Translation:
The force $F_f$ causes acceleration:
$$ma=\mu mg$$
So that:
$$v(t)=\mu gt$$
2. Rotation:
The force $F_f$ also causes a torque $\tau$:
$$\tau=I\alpha,$$
where $\tau=-R \mu mg$, $I=\frac{mR^2}{2}$ and $\alpha=\frac{d\omega}{dt}$, so:
$$-R \mu mg=\frac{mR^2}{2} \frac{d\omega}{dt}$$
$$-2\mu g=R\frac{d\omega}{dt}$$
$$\omega(t)=\omega-\frac{2\mu g}{R}t$$
Rolling without slipping occurs when $v(t)=\omega(t) R$, so with the expressions above:
$$\mu gt=\omega R- 2\mu gt$$
$$t=\frac{\omega R}{3\mu g}$$
Inserting this into $\omega(t)$ we get:
$$\large{\omega(t)=\frac{\omega}{3}}$$
The ring will lose half of its initial rotational momentum before rolling without slipping is achieved.
The initial kinetic energy was:
$$K_0=\frac{mR^2\omega^2}{4}$$
The final kinetic energy $K(t)$, including translational energy is:
$$K(t)=\frac{mR^2}{4}\frac{\omega^2}{9}+\frac{m}{2}\frac{\omega^2 R^2}{9}=\frac{mR^2\omega^2}{12}$$
The equation of motion
$$
\text{torque about stationary geometrical point O} = \text{moment of inertia w.r.t. O} \times \text{angular acceleration w.r.t. O}
$$
is valid only if the motion of the body is planar rotation around an axis that passes through O. This is the case if the point O is taken to be point of contact of the body when rolling without slipping, but not when rolling with slipping. Generally valid version of torque-angular momentum theorem is
$$
\text{torque about stationary geometrical point O} = \ = \frac{d}{dt}\left(\text{angular momentum w.r.t. stationary geometrical point O}\right).
$$
If the body is rolling with slipping, there is no stationary geometrical point O on the ground for which the angular momentum could be written as
$I_O\omega_O$ with $I_O$ constant in time and the latter equation doesn't reduce to the former one.
Best Answer
If there is slipping and friction you now have to deal with an accelerating (non-inertial) frame of reference and axis of rotation.
To use Newton's laws of motion in the non-inertial frame of reference a pseudo-force has to be introduced which acts at the centre of mass, has a magnitude equal to the frictional force and acts in the opposite direction to the frictional force.
That pseudo-force produces a torque about the point of contact with the ground which changes the angular momentum of the disc about that point.