Why is the electronic wavefunction of helium a product of the two 1s wavefunctions when ignoring electron-electron repulsion? This question has also been asked on physics stack exchange:
Why is the helium atom wavefunction a product of the two 1s wavefunctions?

From [1, p. 116]:

In seeking an approximation to the ground state, we might first work out the solution in the absence of the $1/r_{12}$ term. In the Schrodinger equation thus simplified, we can separate the variables $r_{1}$ and $r_{2}$ to reduce the equation to two independent hydrogen like problems. The ground state wavefunction (not normalized) for this hypothetical helium atom would be:
$$\psi(r_1, r_2) = \psi_{1s}(r_1)\psi_{1s}(r_2) = e^{−Z(r_1 + r_2)}$$

Why is it only the product and not some linear combination of the two wavefunctions? I heard somewhere that it has something to do with &quot;tensor product&quot;. Can someone provide a detailed explanation about this?
Reference:

Blinder, S. M. Introduction to Quantum Mechanics: in Chemistry, Materials Science, and Biology; Elsevier, 2012,. ISBN 978-0-08-048928-5.
The same mechanics as for the separation of angular and radial parts of the hydrogen electronic wavefunction are in place here: You have a Schrödinger equation that is separable into two (i.e. one for each electron). You solve them and in order to combine them, they must be multiplied. 

It so happens in this case that the solution for each equation is known and recognizable: the electronic wavefunction of $\ce{He+}$. The fact that the two wavefunctions are multiplied is not related at all to electronic structure theory per se (no involvement of LCAO etc.), but merely a result of the separability of the Schrödinger equation.