One of the deep properties of space that physicists believe to be true is that space is isotropic; that is, in a vacuum one direction is no better than any other.
Isotropy holds equally well if the system in question is spherically symmetric: as long as there is no way to distinguish one direction from the other, you could not possibly expect the physics to behave in different ways in different directions...
This is very good news, because your uniformly charged shell system is spherically symmetric! The charged shell determines a natural origin at the center of the shell, but all directions from this origin are still equivalent, since a spherical shell has no structure that could determine a $``$special$"$ direction. Thus, at a given radius $r$ from the center of the shell, any physical quantity you may want to determine, including the electric field, must be the same in all directions.
(One way to formalize this argument is as follows: assume for contradiction that your uniformly charged shell results in an electric field that is not spherically symmetric. Then there must be some radius $r'$ for which $\vec{E}(r', \theta_1, \phi_1) \neq \vec{E}(r', \theta_2, \phi_2)$; that is, two different directions at the same radius give distinct results. But now rotate the charged shell in space so that $\theta_1 \to \theta_2$ and $\phi_1 \to \phi_2$. 499
Since the charged shell is spherically symmetric, after this rotation nothing will have changed, meaning the electric field at each point in space should be the same as before. But this means that $\vec{E}(r', \theta_1, \phi_1) = \vec{E}(r', \theta_2, \phi_2)$, a contradiction! Thus, the electric field must depend only on $r$.)
With this spherical symmetry in hand, we can now apply Gauss's Law. Since the electric field must be the same for all directions, choosing our Gaussian surface to be a sphere with radius $R$ smaller than the radius of the charged shell, we have $$\unicode{x222F}E\cdot \hat{n} dS = 4\pi R^2E(R) = 0,$$ which of course implies that $E(R) = 0$ for all $R$ enclosed in the uniformly charged shell.
To recap, other field configurations are not possible because they would violate spherical symmetry and/or Gauss's Law. Using symmetries to solve problems in physics is a very powerful, but often quite subtle, skill, and one that is worth developing as soon as possible. 903