In your particular case where you consider separable wave functions (Hamiltonian has your particles non-interacting), I don't get why you say that the form $\psi_{TOTAL} = \frac{1}{\sqrt{2}}[\psi_1(r_1) \psi_2(r_1) - \psi_2(r_1) \psi_1(r_2)]$ is not necessarily a solution.
Indeed, this is just a linear combination of Hartree products of one-electron wavefunctions. You were OK to accept those procuts individually as a solution, so why would a linear combination not be?
Plus, if you do understand that $<\psi_1(r_1) \psi(r_2)| H |\psi_1(r_1) \psi(r_2)> = E_1 + E_2 = E$, you can check very easily what is going on for your slater determinant:
$$ H |\psi_1(r_1) \psi_2(r_2)-\psi_2(r_1) \psi_1(r_2)> = H |\psi_1(r_1) \psi_2(r_2)> -H |\psi_2(r_1) \psi_1(r_2)>$$
$$= (E1+E2)|\psi_1(r_1) \psi_2(r_2)> - (E1+E2)|\psi_2(r_1) \psi_1(r_2)> = E |\psi_1(r_1) \psi_2(r_2)-\psi_2(r_1) \psi_1(r_2)>$$
so that $<\psi_{TOTAL}|H|\psi_{TOTAL}>=E$, giving you the same Energy
EDIT: To see that the exchanged product yields the same value, consider that you can spearate your Hamiltonian in two: $H=H_1+H_2$ where
$$H_i= \frac{-\hbar^2}{2m} \nabla^2_i + V(r_i) $$
each of those hamiltonian acts only on one particle so you have:
$$H |\psi_1(r_1) \psi_2(r_2)> = (H_1 |\psi_1(r_1)>)\otimes|\psi_2(r_2)> + \psi_1(r_1)> \otimes (H_2 |\psi_2(r_2)>) $$
with $H_1 |\psi_1(r_1)>=E_1|\psi_1(r_1)>$ and $H_2 |\psi_2(r_2)>=E_2|\psi_2(r_2)>$
Now, you can realize that the two hamiltonians are the same, only they are acting on different particles. Right? So if you have $H_1 |\psi_1(r_1)>=E_1|\psi_1(r_1)>$, you MUST have $H_2 |\psi_1(r_2)>=E_1|\psi_1(r_2)>$ because H1 and H2 are those same hamiltonians (only acting on r1 or r2), and give the same eigenvalue for the same eigenvector.
Therefore:
$$H|\psi_2(r1)\psi_1(r2)>=(H_1 |\psi_2(r_1)>)\otimes|\psi_1(r_2)> + \psi_2(r_1)> \otimes (H_2 |\psi_1(r_2)>) $$
$$=E_2 |\psi_2(r_1)>\otimes|\psi_1(r_2)> + E_1\psi_2(r_1)> \otimes |\psi_1(r_2)>=(E_1+E_2)|\psi_2(r_1)\psi_1(r_2)>$$