Does a 2D (real-space) wavefunction always has to be a product of two 1D wavefunction (i.e. always separable)?

No, this is not generally true. The reason is that you are assuming that a general two-particle state is given by $|psi_A rangleotimes |psi_Brangle$, and this is not the case. A general two-particle state is a linear combination of product states, i.e.

$$|Psirangle = sum_n c_n |psi_{A_n}rangle otimes |psi_{B_n}rangle $$ In the position basis, we would write

$$|Psirangle = int mathrm dx_1 int mathrm dx_2 psi(x_1,x_2) |x_1rangle otimes |x_2rangle $$

where $psi(x_1,x_2)$ is an element of $L^2(mathbb R)otimes L^2(mathbb R)simeq L^2(mathbb R^2)$ and is under no obligation to factor. If it does factor, so $psi(x_1,x_2) = psi_A(x_1) psi_B(x_2)$, then we can write

$$|psirangle = left(int mathrm dx_1 psi_A (x_1) |x_1rangle right)otimes left(int mathrm dx_2 psi_B(x_2) |x_2rangle right) = |psi_Arangleotimes |psi_Brangle$$

but there is no reason to expect this to be true a priori. In fact, if the two particles are indistinguishable fermions then $psi(x_1,x_2) = -psi(x_2,x_1)$, which is sufficient to show that they cannot be in a product state.

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You may be interested in this answer I wrote about the difference between the direct product $mathcal H_A times mathcal H_B$ and the tensor product $mathcal H_A otimes mathcal H_B$. The key difference is the central point of this answer, namely that elements of the former are all product states while elements of the latter are linear combinations of product states. Modeling composite systems using the latter construction opens up the possibility of having non-product states, which are usually called entangled in a physics context.