Why is there no electric field inside a conducting shell but a gravitational field inside a massive shell? Is there a simple explanation of this observation?
One could even take a liquid mass shell (like a soap bubble) to make the analogy even closer. Why is there such a difference inside the shells between the two cases?
Or: what is the root cause for the difference?
Or otherwise: why don't we float inside a closed elevator?
 A: If I understand your setting correctly, you are asking about a scenario with some outer source of an (electric or gravitational) field, some hollow (conductive or massive) shell, and asking for the resulting field inside this shell.
You even allow for a liquid shell, so the gravity-producing atoms of the shell can move "freely", in analogy to the free-moving charges (eletrons) in a conductor.
The difference is that:

*

*With electric fields, charges of the same sign repulse one another.

*With gravity, masses of the same sign attract one another.

In both cases, the atoms/electrons will move as long as there is a force exterted on them, i.e. as long as the gravity/electric field isn't zero. An equilibrium is only possible when the superposition of the external field plus the field produced by the movable masses/charges sums up to zero.
In the electric case, let's assume the external field is caused by a negative charge "below" the shell. The electrons (negative charges as well) in the shell will experience a repulsion force, making them move upwards. In this direction, they start to compensate the external field, and continue to do so until equilibrium (resulting field = 0) is reached.
In the gravity case, with a mass (e.g. Earth) below the shell, the movable atoms get attracted downwards, which will which will not compensate, but instead increase the field strength, thus moving the system further away from equilibrium.
ADDITION
One more difference between electric fields and gravity is the field magnitude you can achieve in real-world settings:
For most situations, a conductor contains plenty of electrons to compensate the external field.
But if you want to compensate e.g. Earth's gravity field by placing a moderate-size mass close to the "Zero-G" location you want to achieve, you need a very, very dense material. E.g. to achieve a 1g gravity with a 1m diameter massive sphere, you need a material 13,000,000 times denser than Earth, coming close to a neutron star's density.
A: You are mixing up two related ideas.
The absence of a gravitational force inside of a spherical shell of mass was first proved in Newton’s Principia.
Electrostatic attraction and repulsion is a $1/r^2$ force, just like gravity, and so the same proof applies for electrostatics: there is no electrical field inside a uniform spherical shell of charge.
However, a conducting shell is not necessarily a uniform sphere of charge.  Conductors have the property that some charges are free to move about, and will do so (in the electrostatic limit) until there is zero electric field everywhere in the conductor.  In practice this means that all the charges migrate to the conductor’s surface, and therefore that any (charge-free) voids within the conductor will also have zero electric field.
A metal elevator acts as a Faraday cage, blocking external electric fields, because the distribution of charges on its surface is free to be non-uniform.  A spherical shell of mass, on the other hand, does not shield you against gravitational fields from masses outside the shell.  You don’t float in an elevator because the elevator’s mass distribution cannot protect you from the mass of the rest of the Earth.
A: 
Why is there no electric field inside a conducting shell but a gravitational field inside a massive shell?

@JohnRennie commented correctly that:

There is no gravitational field inside a massive spherical shell.

A shell of uniform mass density creates no gravitational field within the shell.
This fact was proven by Newton, and is known as the Shell Theorem. It applies equally to gravitational and electrostatic fields (assuming in the case of an electrostatic field that the charge density at any given distance from the center is uniform.)
If an externally generated field (gravitational or electrostatic) happens to be present, the gravitational and electrostatic cases will behave differently due to the fact that charge can migrate in a conductor, but (solid) mass cannot.
The case of a conductive shell the charge density will be uniform at any given distance from the center if there is no external electric field applied. However, if an external electric field is applied, the charge density will not be uniform. However, according to Gauss's law, and the presuppositions of electro-statics, if there is no charge within the conductive shell, the electric field within the shell will be $0$. Interestingly, this is true for conductors with a cavity within them regardless of their shape. They do not need to be spherical shells. This effect is known as electrical shielding.
In the case of a massive shell with an external gravitational field caused by some external massive object, the following will hold. The massive shell will not generate a gravitational field within the interior of the shell. However, the externally generated gravitational field will penetrate through the shell and be present within the interior of the shell, as if the shell were not there. That is, the massive shell does not shield against externally generated gravitational fields. The reason being that the massive shell will (presumably) not rearrange its mass density the way an conductive shell will rearrange its charge density.
A: The Faraday cage effect of electricity (no electric field inside a conductor) is based on the fact, that charges on a conductor which is subject to an exterior electric field will move until the local surface charge amounts to the normal component of the electric field strength on the surface and the tangential component of the electric field becomes zero. One can show then, that these surface charges exactly compensate the exterior field inside the conductor.
So, the essence of the effect is not so much tied to the fact that conductors for electric charges exist (i.e. fluid masses don't help you with gravity at all), but rather to the fact that charges can always be arranged in such a way (even "by hand", at least conceptionally) that the exterior field gets compensated inside a certain (chargeless) region by the field of the additional charges.
The proof that this is not possible for gravitation need only show that it is already impossible for the special case of a small region where the exterior field is sufficiently homogeneous. In order to compensate the exterior field there, you would need to be able to place gravitational charges (masses) on the boundary of this region, which could also generate a homogenous field of opposite direction. A homogeneous field inside a region will result in a net field flux of zero over the boundary of that region (because the homogeneous field has zero divergence), and hence, a net charge of zero on the boundary.
However, contrary to electricity, there exist only gravitational charges of the same sign (positive masses). The consequence of this is that any mass distribution inside or on the boundary of a certain region will have net gravitational field flux (field lines) that is directed inside the boundary of that region (Gauss' law). This is equivalent to saying that the gravitational charge on the boundary is always nonzero, and hence, cannot generate a homogeneous field.
The example of the spherical mass shell illustrates that there actually exist special mass configurations inside of which gravity is zero. But this is not suitable for compensating an external gravitational field as well, as illustrated above.
A: The only way for a field to be present inside a a conductive shell is to be charge present inside the shell. Because there is no charge present inside the shell, there is no field.
The same holds for mass. If there is no mass inside the shell, there is no gravitational field inside the shell.
The situation is different though if we place them in an external electric or gravitational field.
If we place the massive shell in an external gravitational field, there will be a gravity field inside the shell, equal to the external field.
In the case of an external electric field, the charges in the shell arrange to compensate for the external electric field and the internal field will be zero. The external field is is kept out from a closed conducting shell. The Faraday cage. This can't happen for mass.
Note that in both cases there is no internal charge or mass, but the external fields are not caused by a part of the spheres. The presences of charges outside the shell is considered zero.
For massive spheres the situation without external fields is different.
The gravitational field decreases linearly towards the center of the sphere, from the surface. There is mass present inside each closed surface inside the sphere creating a force towards the center. The gravity on the surface of a sphere with constant density and decreasing R decreases goes down linearly with R. The gravity experienced inside the Earth about 3000 km. deep is about half as that experienced on the surface. The shell around it doesn't contribute.
In the case of the conducting sphere, the presence of surface charges causes an internal field that is zero. Only outside the sphere there is a field. If a field would be present it would move the charges until the field is gone.
