The orbital wavefunctions of the hydrogen atom, which obey the eigenvalue equation
$$
\left[-\frac{1}{2\mu}\nabla^2-\frac{e^2}{r^2}\right]\psi_{nlm}=E_{nl}\psi_{nlm},
$$
are functions of the separation vector $\mathbf r$ which points from the proton towards the electron. This is a standard trick in the two-body problem and it is done in both the classical and the quantum versions to factor away the motion of the bigger body (which is close to the centre of mass) and leave an effective one-body problem which is easier to treat.
This means that the orbital angular momentum, with total angular momentum number $l$, is in fact the combined angular momentum of the electron and the proton about their centre of mass. In essence, the proton partakes in part of the orbital motion and takes out some of the angular momentum from the electron. (Note, though, that this is classical language and it explicitly does not hold for the hydrogen atom, where the angular momentum of the motion is essentially indivisible.)
This raises an apparent paradox, which is resolved through the fact that the separation vector obeys dynamics through the reduced mass $\mu=1/\left(\tfrac{1}{m_e}+\tfrac{1}{m_p}\right)\approx\left(1-O\left({m_e\over m_p}\right)\right)m_e\lesssim m_e$ of the system, and this is slightly smaller than the electron mass. This slightly enlarges the orbital radius of the electron (since the Bohr radius is inversely proportional to the mass). The velocity stays constant (at $\alpha c$), which means that the angular momentum $L\sim \mu r v$ stays constant as well.
That said, the proton does have spin angular momentum of its own, but this couples weakly to the electronic motion. This coupling is via the same spin-orbit couplings as the electron, but its much higher moment of inertia means that the relevant energies are much smaller, as are the corresponding hyperfine splittings in the spectrum.
For a full shell, the addition of the expectation values of any angular momentum $L_i$ is zero, and similarly for the spin operators $\sigma_i$. This is not hard to see - for $l(l+1)$ as the expectation value of $L^2$ for a s,p,d,f subshell, the basis of that subshell is spanned by states indexed by integers between $-l$ and $l$, and since that l is also the actual value for the angular momentum in a given direction, full occupancy of these states implies vanishing total angular momentum. Even easier for spin - there are only up and down states, and equally many of them in every shell, so if they're all occupied, the net spin is zero.
Best Answer
We describe the whole system with a state, this state is a combination of the single particle states (orbitals). Each orbital we define in terms of an orbital momentum shell. A full shell has zero total angular momentum, therefore multiple full shells still have zero total angular momentum. Finally a full shell combined with a few valence electrons in higher orbitals would have the angular momentum of only the valence electrons. Now I will demonstrate why a full shell must have zero angular momentum.
An example using the simplest S-shell.
We have two states available, "up" $|\uparrow{} \rangle$, and "down" $|\downarrow{} \rangle$. We also have the constraint that these are fermions, meaning any combination has to be entirely antisymmetric when two particles are interchanged.
If we are placing a single electron into the S-shell we have 2 states available, either: $$|\uparrow{}\rangle \text{ & } |\downarrow{}\rangle$$ Each of these have angular momentum $\frac{1}{2}$. However if we want to add another electron we only have 1 possible state which satisfies antisymmetry, $$|\psi\rangle = \frac{1}{\sqrt{2}}\left(|\uparrow{}\downarrow{}\rangle - |\downarrow{}\uparrow{}\rangle\right)$$ This is the singlet configuration which has angular total angular momentum zero.
Importantly there is only one state with total angular momentum $J=0$, two states with $J=\frac{1}{2}$, three states with $J=1$ (triplet), and so on.
Here I will outline the logic of the general proof. Firstly ignoring spin, each single particle orbital $l$ has $2l +1$ states with angular momentum projections ranging from $-l \leq m_l \leq l$. Secondly not ignoring spin you can place 2 electrons in each orbital with spin up or down. This gives $2(2l + 1)$ single particle states. If we have completely filled this shell that means that we have placed an electron in each single particle orbital.
Now if we count all of the unique ways we can fill all of the orbitals, there is only one way to do this. That means that this is a singlet configuration and not a member of a higher multiplet (such as the triplet with 3 states mentioned above).
The total state is defined in terms of having a total angular momentum and total angular momentum projection. Clearly it has angular momentum projection of $0$ since $\sum_{m= -l}^{m = l} m = 0$. However since it is a singlet state it also has total angular momentum $0$, and we can treat it like a "core" with no angular momentum.
As a side note this is used extensively in atomic physics as well as nuclear physics. For nuclear physics we would not be talking about electrons but instead protons and neutrons. Therefore we not only have the choice of spin up or down for each orbital but also between proton and neutron. This gives us 4 particles in each orbital, and is made more rigorous with the idea of "isospin". So far as most nuclear interactions care their interactions are the same so we can treat them as 2 projections of one object, the "nucleon". The total wavefunction must be antisymmetric under interchange in the combined spatial-spin-isospin space. A filled orbital momentum shell would therefor have zero angular momentum as well as zero total isospin.