What will happen to a ball kept on a frictionless inclined plane?

...the torque exerted by $N$ is zero but the torque exerted by $mg$ is non-zero. This means the ball must roll...

Actually, it means that the angular momentum about that axis must increase. That is not the same as rolling. If the axis is through the center of mass of the object then the only way for the angular momentum to increase is through rolling. However, if the axis does not pass through the center of mass then there is also angular momentum due to the linear motion. In other situations this is the difference between orbital angular momentum and spin angular momentum. So let's calculate the "orbital" angular momentum in this problem.

The torque is $m g R \sin(\theta)$ where $R$ is the radius of the ball and $\theta$ is the angle of the incline.

The magnitude of the "orbital" angular momentum is given by $R m v$ where $v$ is the linear velocity of the center of mass, so its time derivative is $R m a$ where $a$ is the linear acceleration of the center of mass.

From Newton's laws the linear acceleration is the component of gravity which is down the slope. This is $ma=mg \sin(\theta)$ so $a=g \sin(\theta)$.

Substituting the linear acceleration into the time derivative of the orbital angular momentum gives $R m g \sin(\theta)$ which is equal to the torque. This means that the increase in angular momentum due to the torque is fully accounted for by the increase in the "orbital" angular momentum and there is no left over torque for increasing the "spin" angular momentum. Therefore, the ball does not spin/roll regardless of which axis you examine.

So, what exactly will happen to a ball kept on a frictionless inclined plane - will it slide or roll?

Frictionless means the surface of the incline cannot exert any torque on the ball. By Newton's second law, that means the state of rotation of the ball remains unaltered, specifically:

• if the ball was spinning at angular velocity $\omega$ then it will simply continue to do so: $\frac{\text{d}\omega}{\text{d}t}=0$.
• if the ball wasn't spinning at all ($\omega=0$) then sliding down the frictionless incline will not alter $\omega$ . Again $\frac{\text{d}\omega}{\text{d}t}=0$.

For any change in rotational status to occur, a torque $\tau$ needs to act on the ball, so that:

$$\tau=N\mu$$

but with $\mu=0$, $\tau$ is always $0$.

to see what happened, lets look at the equations of motion:

$$m\,\ddot{s}+F_c-m\,g\sin(\alpha)=0\tag 1$$ $$I_b\,\ddot{\varphi}-F_c\,R=0\tag 2$$

case I: Ball is rolling without slipping:

$$\ddot{s}=R\ddot{\varphi}\tag 3$$

you have three equations for three unknowns $\ddot{s}\,,\ddot{\varphi}\,,F_c$

you obtain:

$$\ddot{\varphi}=\frac{m\,g\,\sin(\alpha)\,R}{m\,R^2+I_b}$$ $$\ddot{s}=R\ddot{\varphi}$$ $$F_c=\frac{I_b\,m\,g\,\sin(\alpha)}{m\,R^2+I_b}$$

case II: Ball is sliding :

This is your case, because you don't have the contact force $F_c$.

in this case the contact force $F_c$ is equal zero.

$$m\,\ddot{s}=m\,g\sin(\alpha)$$ $$I_b\,\ddot{\varphi}=0\quad \Rightarrow \varphi=0$$

case III: Ball is rolling :

instead of equation (3) you have now the equation for a friction force

$$F_c=\mu\,N=\mu\,m\,g\,\cos(\alpha)$$

you obtain:

$$\ddot{s}=g(\sin(\alpha)-\mu\,\cos(\alpha))$$

$$\ddot{\varphi}=\frac{\mu\,m\,g\,\cos(\alpha)\,R}{I_b}$$

so if $\mu=0$ the ball is sliding this is case II