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 formula you mention is for general motion of an object that is both rotating and translating. Let $\mathbf{r}$ be the position of a point in the object and $\mathbf{r}_{\text{CM}}$ be the position of the center of mass of the object. We define the vector
$$\mathbf{r}' \equiv \mathbf{r} - \mathbf{r}_{\text{CM}}$$
which represents the position of a point relative to the center of mass. The angular momentum $\mathbf{L}$ can be found by
$$\mathbf{L} = \int \mathbf{r} \times \text{d}\mathbf{p} = \int \mathbf{r} \times \dot{\mathbf{r}} \text{d}m = \int \left(\mathbf{r}_{\text{CM}}+\mathbf{r}'\right) \times \left(\dot{\mathbf{r}}_{\text{CM}}+\dot{\mathbf{r}}'\right) \text{d}m \\ = \int \mathbf{r}_{\text{CM}} \times \dot{\mathbf{r}}_{\text{CM}} \text{d}m + \int \mathbf{r}' \times \dot{\mathbf{r}}'\text{d}m + \int \mathbf{r}_{\text{CM}} \times \dot{\mathbf{r}}'\text{d}m + \int \mathbf{r}' \times \dot{\mathbf{r}}_{\text{CM}}\text{d}m$$
The last two terms vanish by the definition of the center of mass $\int \mathbf{r}' \text{d}m = 0$, so we have
$$\mathbf{L} = \int \mathbf{r}_{\text{CM}} \times \dot{\mathbf{r}}_{\text{CM}} \text{d}m + \int \mathbf{r}' \times \dot{\mathbf{r}}'\text{d}m \\ = m\mathbf{r}_{\text{CM}} \times \dot{\mathbf{r}}_{\text{CM}} + \int \mathbf{r}' \times \left(\boldsymbol{\omega} \times \mathbf{r}'\right)\text{d}m \\ = \mathbf{r}_{\text{CM}} \times \mathbf{p}_{\text{CM}}+ I\boldsymbol{\omega}$$
The first term is the $mvr$ that you mention which is commonly called the "orbital" angular momentum. The second term represents the rotation (spin) of the body.
The point is that the first term only makes sense when you consider an origin. In this case, the origin can be taken as any point fixed on the surface. The center of mass then moves in a straight line above the surface and parallel to it. You can check that $\mathbf{r}_{\text{CM}} \times \mathbf{p}_{\text{CM}}$ is constant.
The body does not have to actually orbit the origin. As long as the position and velocity of the center of mass are not parallel, the first term will be non-zero. It will only be zero if the body is heading directly towards or away from the origin. In other words, it can be thought of as how much the body is "going past" or "missing" the origin.
Best Answer
Your solution looks fine to me.
Yes: the angular momentum is preserved in the horizontal plane (the weight is vertical and the reaction of the sphere surface is a central force) so your first relation is fine, just remember that $\theta$ is not the vertical angle, but lies on the plane tangent to the sphere at point B.
There are two kinds of rotational energy: the one of the particle spinning on itself which would require the particle's mass distribution to be computed, which is not given so I assume it should be neglected. What may be confusing is the particle's rotational energy around the centre of the sphere: $$E=\frac{1}{2}I\omega^2$$ For a point like mass the moment of inertia is: $$I = mr^2$$ Remembering the relation between angular and tangential velocity: $$\omega= \frac{v}{r}$$ we get exactely: $$E=\frac{1}{2}mv^2$$ In fact for a point like mass it is exactly the same to consider the kinetic energy related to the tangent velocity, or the rotational kinetic energy related to the angular velocity, for this problem I approve your choice for the first one.