Which expectation values can be determined with KS orbitals?

The general answer is: no. Take the total spin, $\hat{\bf S}^2$, for example. Although quantum chemistry programs typically compute the expectation value $\langle \hat{\bf S}^2 \rangle$ using the non-interacting orbitals similarly to Hartree-Fock, this procedure is not founded in theory. For more details, I refer to Jacob and Reiher's paper on Spin in Density-Functional Theory.

Since you ask "which" expectation values can be determined this way, instead of "can all" expectation values be determined this way, I'll briefly mention one case where the Kohn-Sham orbitals indeed result in the correct expectation value.

Consider the case where $A=A(r)$ is a local operator (which is necessarily a one-electron operator). Then $$ \langle \Psi, A\Psi \rangle = n\int A(r_1)\Psi^*(r_1,\ldots,r_n)\Psi(r_1,\ldots,r_n) dr_1 dr_2 \cdots dr_n = \int A(r)\rho(r) dr \tag{1} $$ where $n$ is the number of electrons and $\rho(r)$ is the density, since $$ n\int \Psi^*(r_1,\ldots,r_n)\Psi(r_1,\ldots,r_n) dr_2 \cdots dr_n = \rho(r_1) \tag{2} $$ On the other hand, for the Kohn-Sham determinant we have $$ \langle \psi_1\cdots\psi_n, A\psi_1\cdots\psi_n \rangle = \sum_{i=1}^n\int A(r)\psi_i^*(r)\psi_i(r) dr = \int A(r)\rho(r) dr \tag{3} $$ since the density built from the Kohn-Sham orbitals is equal to the exact density: $$ \sum_{i=1}^n \psi_i^*(r)\psi_i(r) = \rho(r) \tag{4} $$ This proves that the expectation value of a local operator is exactly given as its expectation value over the Kohn-Sham determinant.

Meanwhile, not all one-electron operators have this property. For example the expectation value of the kinetic energy operator over the Kohn-Sham determinant gives the non-interacting kinetic energy $T_s$, which is famously not equal to the exact kinetic energy (although usually very close to it - that's why Kohn-Sham DFT works so much better than orbital-free DFT with common approximate functionals); part of the kinetic energy is in the exchange-correlation energy, specifically the correlation part. Two-electron operators in general do not have this property either, as demonstrated by Susi's nice example. It is however an interesting question whether there are non-local operators whose expectation value over the Kohn-Sham determinant always agrees with the expectation value over the many-electron wavefunction $\Psi$, for which I have no answer...

It cannot hold for all observables. This is a more or less straightforward consequence of the Hilbert space structure of quantum mechanics and DFT plays no role.

Indeed, consider two normalized vectors $\psi,\phi\in H$. If for all hermitian $A\in B(H)$ it holds that $\langle \psi,A\psi\rangle=\langle \phi,A\phi\rangle$, then $\psi=e^{i\varphi}\phi$ for some $\varphi\in \mathbb R$.

Hence, if $\psi$ and $\phi$ correspond to a different state (i.e. differ by more than a phase), then there is at least one (bounded) observable which recognizes this.

As an explicit and easy example, consider the observable $A=|\psi\rangle\langle\psi|$. Then

$$\langle \phi,A\phi\rangle=|\langle \phi,\psi\rangle|^2< ||\psi||^2 ||\phi||^2=1=\langle \psi,A\psi\rangle\quad , $$

by Cauchy-Schwarz, as long as $\psi$ and $\phi$ differ by more than a phase.

Thus, as long as the many-body state $\psi$ is not equal to the "Kohn-Sham state" $\phi=\psi_1\cdots \psi_n$ there always exist observables which can distinguish these states by their expectation values.