Electric field in a shell in asymmetric situation

Consider a conductor of any shape filled inside with a conducting material and having charge Q on it. Now bring a charge Q1 near it. The charges on the outer surface of conductor will align so that field inside is zero everywhere. Now after the electrostatic condition has been established, I scoop out a cavity inside the conductor. Assuming perfect vacuum, notice that the forces on any of the charges on the conductor or on the charge Q1 do not change. This means that the charge distribution is unchanged after scooping out cavity. So even after scooping out cavity, the field inside the cavity is still zero everywhere. Now I can make this cavity large enough so that the remaining part is just a shell and this argument will still be valid. So in your case also, field inside shell is zero everywhere. Now by the reasoning it is clear why this will not necessarily hold in the case of a non conducting material.

This result can be generalized further, although the rigorous proof is through the Laplace equation. Consider a conducting shell( any shape) with a cavity having some charge (which implies there is equal and opposite charge on the inner surface of cavity) and there is also some charge outside the conductor and on the outer surface of the conductor. We find that net field due to charge on outer surface and due to charge outside the shell is zero everywhere inside the cavity and inside the conducting material. Also net field due to charge inside the cavity and the charge on the surface of cavity is zero everywhere outside the cavity. The conductor sort of acts as a shield against the 2 fields. This phenomenon is called electrostatic shielding