This question is an extension of a question I asked earlier about the possibility of measuring the energy of just one particle in a system of many identical particles, which can be found here. One answer confirmed what I already suspected in my question, namely that one cannot measure the energy of a single-particle constituent of an many-particle system. I assume this is because the eigenstates of the single-particle operators are not acceptable correctly-symmetrised many-particle wavefunctions.
If we now extend the formalism of my earlier question to other observables (and thus other Hermitian operators), I am led to believe that one cannot measure an observable of just one particle in a system of many identical particles. Is this correct? In particular, consider the case of a system of two particles (non-identical or identical) in the spin-singlet state: $$\lvert\psi\rangle = \frac{1}{\sqrt{2}}(\lvert\uparrow\rangle_{1} \lvert\downarrow\rangle_{2} - \lvert\downarrow\rangle_{1} \lvert\uparrow\rangle_{2})$$. In Bohm's version of the EPR argument, a system of an electron and a positron (which are non-identical) starts out in this state. Then a measurement of $S_{z}$ is performed on the electron and it is, for example found in state $\lvert\uparrow\rangle_{1}$. The state of the two particles after the measurement is thus $\lvert\psi'\rangle = \lvert\uparrow\rangle_{1} \lvert\downarrow\rangle_{2}$, which is an acceptable state for two non-identical particles. Therefore, the state of the positron has been affected by a measurement on the electron.
However, if we now take two electrons and prepare them in the singlet state, this reasoning can no longer apply. This is because, as mentioned above, one cannot measure the spin of just one of the two identical electrons and $\lvert\psi'\rangle = \lvert\uparrow\rangle_{1} \lvert\downarrow\rangle_{2}$ is not an acceptable wavefunction for two identical particles. Does this mean that one cannot perform the EPR experiment with two identical particles?