Why don't electrons in $s$ orbitals scatter off the nucleus? Electrons in $s$ orbitals statistically spend the better part of their time in the nucleus. What stops them from scattering off the nucleus?
 A: I think the comments may be spending too much time on semantics instead of addressing what I think is the conceptual point of your question.
The notion of scattering or colliding in the sense of two particles coming in, bouncing off each other, and then proceeding in deflected directions, is a very classical one. In that the interaction is very localised in space and in time: the particles do not feel each other's presence while incoming or outgoing, but only interact when their surfaces meet, at a specific point in space and at a specific instant in time.
Quantum particles such as electrons are described by wavefunctions, which are delocalised in space and hence not really capable of performing the sort of collisions and scattering processes described in the previous paragraph. The "presence" of the electron and the nucleus are felt by each other, in principle, even at infinity, since the wavefunctions decay exponentially (usually).
The crucial point to mention is the wavefunction for electrons in $s$ orbitals (or any orbitals) are bound states - bound, not free. Meaning that you obtained them from the Schrödinger equation with the Coulomb potential of the nucleus already present! That is, the trajectory of the electron in the $s$ orbital, by definition, already takes into account of the "interaction" (or "presence of") the nucleus.
The "simple" $s$ orbital wavefunction is usually derived from the $1/r$ Coulomb potential which hence treats the nucleus as a point particle. This is not very physical as the nucleus has a spatial extent, which causes both a quadrupole moment in its electric field (taken into account by the hyperfine structure) and a smearing of the electrostatic interaction around the nucleus (taken into account by the Darwin term). For $s$ electrons, only the Darwin effect matters.
On top of that, while rare, the electrons can interact with the protons in the nucleus via electron capture: a proton "absorbs" an electron and turns into a neutron. This, for example, happens during the collapse of some stars, where the gravitational pull raises the electron Fermi energy and makes electron capture energetically favourable.
If you really want to find an analogous "scattering" process with electrons and the nucleus, then you have to take free electrons, and shoot them at the nucleus. They will still have to obey linear and angular momentum conservation, but they will only feel the Coulomb repulsion off the nucleus as a transient ("localised" in both space and time) effect. And this is actually how the nucleus was discovered in Rutherford scattering experiments.
EDIT:
Let me also add that:

*

*a bound electron is a stationary state, which is a consequence of the Hamiltonian being time-independent and which means the probability density distribution does not change with time.


*an $\ell = s$ electron has its probability current equal to $0$.
So take into account these two points while pondering on the philosophical question of whether or not electrons actually "move" inside the atom.
