There is a torque, but it points sideways, perpendicular to the orientation of the hammer at any time. Because of this, as the hammer rotates, the direction of the torque also rotates around with it. The torque only acts to rotate the system horizontally around in space, not to change the direction of its angular velocity.
Let's see this with a calculation. Suppose I model the hammer as a rod of length $L$ and mass $m_r$ with a point mass $m_p$ on the end.
The moment of inertia of the hammer at this moment can be computed by taking the moment of inertia of a similar configuration aligned along the $x$ axis and rotating it by an angle $-\theta$ in the $y$ direction:
$$\begin{align}I &= R_y^{-1}\begin{pmatrix}0 & 0 & 0 \\ 0 & \frac{mL^2}{3} + m_pL^2 & 0 \\ 0 & 0 & \frac{mL^2}{3} + m_pL^2\end{pmatrix}R_y \\
&= \begin{pmatrix}ML^2\sin^2\theta & 0 & -ML^2\sin\theta\cos\theta \\ 0 & ML^2 & 0 \\ -ML^2\sin\theta\cos\theta & 0 & ML^2\cos^2\theta\end{pmatrix}\end{align}$$
where $R_y$ is the rotation matrix around the $y$ axis and $M = \frac{m_r}{3} + m_p$. Computing the angular momentum using $\vec{L} = I\vec\omega$, where $\vec\omega = \omega\hat{z}$, I get
$$\vec{L} = ML^2\cos\theta(\hat{z}\cos\theta - \hat{x}\sin\theta)$$
The torque, on the other hand, is
$$\vec\tau = \vec{r}\times\vec{F} = (\hat{x}\cos\theta - \hat{z}\sin\theta)\times (-mg\hat{z}) = mg\hat{y}\cos\theta$$
So the torque actually pushes the angular momentum in a direction perpendicular to both its direction and the orientation of the hammer. This means there will be no change in the amount of angular momentum. It also means that the hammer rotates along with everything else, so that the angular momentum and the momenta of inertia preserve their relative orientation and thus there will be no change in its angular velocity.
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:
$$K(t)=\frac{mR^2}{4}\frac{\omega^2}{9}+\frac{m}{2}\frac{\omega^2 R^2}{9}=\frac{mR^2\omega^2}{12}$$
Best Answer
The concept of angular momentum only makes sense when we specify a rotation axis. So we will pick the axis passing through the initial center of rotation.
When the string is cut, the point mass has a linear momentum $p = mv = mr\omega$. One can define the angular momentum of a particle about an axis as $\vec{L} = \vec{d} \times \vec{p}$ where $\vec{d}$ is the vector from the axis of rotation to the particle (I used $d$ to avoid confusion with the $r$ you specified). The magnitude of angular momentum is then $L = dmr\omega \sin\theta$, but $\sin\theta$ can be shown from geometry to be $r/d$.
Thus the solution is $L = dmr\omega \frac{r}{d} = mr^2 \omega$, which does not depend on the distance the point particle is from the axis and is the same as the initial angular momentum. Hence angular momentum is conserved.
Now as for where the linear momentum came from. The object always had linear momentum of magnitude $p = mv = mr\omega$, however if you remember, $\vec{F} = \frac{d\vec{p}}{dt}$, and since there was a tension force on the point mass while it was rotating the direction of the momentum was changing. When you cut the string the direction is chosen and remains constant, and linear momentum is conserved.