Acceleration due to a force applied outside the center of mass

Newtons second law $F_{net} = ma$ holds true for a system of particles irrespective of the point of application of force.

"If we were to apply the same force to any other point of the ball, we would then obtain the same acceleration?"

Yes, doesn't matter if it's a rigid body or not, the acceleration of the center of mass of ANY system of particles is $vec{a_{CoM}}=frac{vec{F_{ext}}}{m}$, where $vec{F_{ext}}$ is the sum of external forces acting on ANY particles of the system. This idea is the whole point of the center of mass. You should look at the derivation of the CoM formula in your book. This idea is the exact thing that they prove.

The external force need not be acting on the center of mass itself for this to be true. As center of mass is in many cases an imaginary point where there are no actual particles, it doesn't even make sense for a force to be directly applied on the CoM in the general case of a system of particles.

The reason that $vec{F} = m , vec{a}_{rm com}$ works regardless of where $vec{F}$ is applied has to do with how momentum is defined, and Newton's 2nd law.

Take a system of particles that move and rotate together and you will find that the total combined momentum is

$$ vec{p} = sum_i m_i vec{v}_i = m , vec{v}_{rm com} $$

That is the sum of all individual momenta (particle mass times particle velocity) equals the total mass time the velocity of the center of mass. Note that the momentum is defined for the entire body and not of a particular point on the body.

Now force is the time derivative of momentum, or $vec{F} = frac{rm d}{{rm d}t} vec{p}$. And since $vec{p}$ is defined for the entire body, so is $vec{F}$. That is the force is the net force applied on the body as in

$$ vec{F} = sum_i vec{F}_i $$

The net force is the combined force applied on all the points and it only describes the motion of the center of mass.

$$ vec{F} = m, vec{a}_{rm com} $$