Why is the electronic wavefunction of helium a product of the two 1s wavefunctions when ignoring electron-electron repulsion?
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.