I think this is like the sphere or cylinder rolling on a plane. At each instant the sphere/cylinder is effectively rotating about the instantaneous point of contact, even though this point itself is moving. Same here : at each instant the ball on the string is rotating about the instantaneous point of contact. So angular momentum is constant about the contact point.
When the string has moved through an angle of $\theta$, the contact point has also moved through an angle of $\theta$ around the circle of radius $a$. So the string has shortened from $r_0$ by a distance of $a\theta$.
Therefore at any instant $(r_0-a\theta)\omega=r_0\omega_0$.