It depends on exactly how the question is phrased and interpreted.
If you phrase the question so that the treadmill speed is constantly adjusted to eliminate any forward motion of the plane as seen by an outside observer, then you are also requiring that the plane's acceleration must be zero as seen by an outside observer. This means that you are requiring that the plane's wheels spin fast enough to create a drag force (due to tire rolling resistance and wheel bearing friction) equal to engine thrust, even though they are mounted on the axles on low-friction bearings. That treadmill would have to spin awfully fast to stop the plane from moving forward!
So if we say that "the treadmill turns as fast as the wheels are turning", which is another way of saying "the treadmill moves as fast as the airplane is moving relative to the treadmill", then we are specifying zero ground speed as seen by an outside observer, and the plane can't take off, unless it is extremely windy. But as noted above, this would involve no ordinary treadmill. It might have to spin at thousands of miles per hour to create sufficient drag to counteract thrust!
On the other hand, in the question above, we read:
"it acts like a treadmill moving backwards at the same speed as the
aircraft is going forward"
The key question here is: "going forward relative to what reference frame"?
Taken literally, the most plausible interpretation of the language above is that the speed of the treadmill is equal in magnitude (and opposite in direction) to the groundspeed of the plane as seen by an outside observer. Now we are no longer requiring that the groundspeed of the plane be zero. In fact, we are requiring that the groundspeed of the plane NOT be zero, unless the treadmill is stationary. And we're also requiring that the plane's wheels spin twice as fast as usual at the moment of takeoff. In which the case the plane will probably be able to take off just fine.
(Another answer notes, likely correctly, that many airliner tires might not tolerate double the normal rotation speed; many light plane tires likely would.)
Note that if we modified the language of the question to reference the "speed" displayed on a speedometer connected to the wheels-- not something we normally find on an airplane!-- then we are back to the first interpretation given above, and the plane cannot take off or even move forward at all.
It's also worth noting that if the engine power were actually applied directly to the wheels, the moving conveyor belt would greatly limit the thrust available for a given total power output, and we'd no longer need to spin the belt at a such a ridiculously extreme speed in order to stop the vehicle from moving forward. The belt speed would still have to be well above the vehicle's normal top speed, since aerodynamic drag is not a factor- assuming that the moving belt isn't dragging enough air with it to create a significant headwind!
ALTERNATIVE ANSWER -- as the belt spins at the very high rate of speed required by the first interpretation above, it will "drag the air with it" and create a built-in headwind. The resulting air drag will reduce the total belt speed needed to stop the plane from moving forward. In fact the built-in headwind just might be strong enough to allow the aircraft to take off after all, with zero groundspeed! In this case things will get a bit "interesting" as the plane tries to climb up through the wind gradient!
(All other versions of this question appear to not be accepting new answers, and may not actually be exact duplicates.)