According to Sakurai the solutions of the two-electron system are of the form $\psi=\phi({\bf x_1},{\bf x_2})\chi(m_{s1},m_{s2})$
Since it's a fermionic system, $\psi$ must be a linear combination of antisymmetric states. If $\phi$ is symmetric and $\chi$ is antisymmetric (or the other way around), then $\phi\chi$ is antisymmetric, and so is a linear combination.
With no spin dependence, the Hamiltonian is $\mathcal{H}=({\bf p_1}^2 + {\bf p_2}^2)/2m$, and the spatial solutions are of the form $\omega_A({\bf x_1})\omega_B({\bf x_2})$, so $\phi$ can be written as a symmetrical and antisymmetrical combination
\begin{equation} \phi_{\pm}({\bf x_1},{\bf x_2}) = \frac{1}{\sqrt{2}} \left[ \omega_A({\bf x_1})\omega_B({\bf x_2}) \pm \omega_A({\bf x_2})\omega_B({\bf x_1}) \right] \end{equation}
In the same way, $\chi$ can be a triplet or a singlet state.
But, is every possible solution a linear combination of antisymmetric terms $\phi\chi$? I don't think so, because I found the following state
\begin{equation} \psi = \omega_A({\bf x_1})\omega_B({\bf x_2})\chi_{+-} - \omega_A({\bf x_2})\omega_B({\bf x_1})\chi_{-+} \end{equation}
And I couldn't write it as a linear combination of the following:
\begin{equation} \left\lbrace \begin{array}[l] &\phi_+({\bf x_1},{\bf x_2})\frac{1}{\sqrt{2}}\left( \chi_{+-}-\chi_{-+} \right)\\ \phi_-({\bf x_1},{\bf x_2}) \left\lbrace \begin{array}[l] &\chi_{++}\\ \frac{1}{\sqrt{2}}\left( \chi_{+-}+\chi_{-+} \right)\\ \chi_{--} \end{array} \right. \end{array} \right. \end{equation}
The state $\psi$ is antisymmetric, and it is a valid state for the 2-electron system. But it isn't a combination of antisymmetric states of the form $\phi({\bf x_1},{\bf x_2})\chi(m_{s1},m_{s2})$, so these states do not form a complete basis of solutions. I would like to know a complete basis for the system.
