I know this has been discussed several times, I try to be short and straight to the point.
If we consider in lab frame one ball at rest and the other with velocity $\vec{v_2}$ we have, thanks to conservation of momentum and energy: $$\vec{v_2}=\vec{v_1 '}+\vec{v_2 '}$$ and $${v_2}^2={v_1 '}^2+{v_2 '}^2$$ that directly leads to $$\vec{v_1 '}\cdot\vec{v_2 '}=0$$
This is compatible both with one ball stopping and with 90 degrees scattering.
What is the right answer?
In two or more dimensions, the conservation of total kinetic energy and total momentum alone does not result in a unique solution as there are fewer equations than unknowns. More information, such as the impact parameter, is needed to obtain a unique solution. Both situations you describe are possible.
An impact parameter of zero corresponds to a head-on collision. In this special case, the collision is effectively one-dimensional. Since the two balls are identical, the moving ball transfers all of its kinetic energy and momentum to the other ball and itself becomes stationary.
Otherwise, if the impact parameter is larger than zero and less than the diameter of the ball, then neither resultant velocity is zero. The result you derived implies that they must be perpendicular to each other.
A more general method is to use the straight line passing through the centers of the balls at the instant of collision as the horizontal axis and consider components parallel and perpendicular to it. This is described in this post.
The $90^\circ$ scattering is when the impact parameter is not zero, ie it is an off-centre collision and the other is for a head-on collision.
