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The total wavefunction of an electron $psi(vecr,s)$ can always be written as $$psi(vecr,s)=phi(vecr)zeta_s,m_s$$ where $phi(vecr)$ is the space part and $zeta_s,m_s$ is the spin part of the total wavefunction $psi(vecr,s)$. In my notation, $s=1/2, m_s=pm 1/2$.

Question 1 Is the above statement true? I am asking about any wavefunction here. Not only about energy eigenfunctions.

Now imagine a system of two electrons. Even without any knowledge about the Hamiltonian of the system, the overall wavefunction $psi(vecr_1,vecr_2;s_1,s_2)$ is antisymmetric. I think (I have this impression) under this general conditions, it is not possible to decompose $psi(vecr_1,vecr_2;s_1,s_2)$ into a product of a space part and spin part. However, if the Hamiltonian is spin-independent, only then can we do such a decomposition into space part and spin part.

Question 2 Can someone properly argue that how this is so? Please mention about any wavefunction of the system and about energy eigenfunctions.

Your claim

[any arbitrary] wavefunction of an electron $psi(vecr,s)$ can always be written as $$psi(vecr,s)=phi(vecr)zeta_s,m_s tag 1$$ where $phi(vecr)$ is the space part and $zeta_s,m_s$ is the spin part of the total wavefunction $psi(vecr,s)$

is false. It is perfectly possible to produce wavefunctions which cannot be written in that separable form - for a simple example, just take two orthogonal spatial wavefunctions, $phi_1$ and $phi_2$, and two orthogonal spin states, $zeta_1$ and $zeta_2$, and define
$$
psi = frac1sqrt2bigg[phi_1zeta_1+phi_2zeta_2 bigg].
$$

Moreover, to be clear: the hamiltonian of a system has absolutely no effect on the allowed wavefunctions for that system. The only thing that depends on the hamiltonian is the energy eigenstates.

The result you want is the following:

If the hamiltonian is separable into spatial and spin components as $$ H = H_mathrmspaceotimes mathbb I+ mathbb I otimes H_mathrmspin,$$ with $H_mathrmspaceotimes mathbb I$ commuting with all spin operators and $mathbb I otimes H_mathrmspin$ commuting with all space operators, then there exists an eigenbasis for $H$ of the separable form $(1)$.

To build that eigenbasis, simply diagonalize $H_mathrmspace$ and $H_mathrmspin$ independently, and form tensor products of their eigenstates. (Note also that the quantifiers here are crucial, particularly the "If" in the hypotheses and the "there exists" in the results.)