If the particles never did interact in the past and won't ever interact in the future, then it indeed makes not much sense to model the system as a tensor product. Keep in mind that it does not matter how long ago the interaction happened as long as the particles are not otherwise disturbed - that's one of the points of the EPR 'paradoxon'.
However, care must be taken when dealing with identical particles: They are excitations of the same underlying field and should always be considered as having interacted.
In fact, the state space of identical particles is not the tensor product, but its symmetrization (in case of bosons) or anti-symmetrization (in case of fermions), ie only entangled states are considered.
These are just my thoughts and by no means authorative - anyone who deals with such issues on a more hands-on basis probably has a better grasp of the underlying issuess...
Considering the unexplained downvote, a more technical answer to the problem:
In non-relativistic Quantum mechanics, the evolution of a system in governed by inital state and Schrödinger's equation.
If we start out in an un-entagled state and the Hamiltonian does not contain interactions between the different sub-systems, we will stay in an unentangled state and we can consider the sytems independantly.
More formally, let our inital state be
$$\psi = \psi_1\otimes...\otimes\psi_n$$
and the Hamiltonian given by
$$H = \tilde H_1 + ... + \tilde H_n$$
$$= H_1\otimes1\otimes...\otimes1 + 1\otimes H_2\otimes1\otimes....\otimes1 + ... + 1\otimes...\otimes1\otimes H_n$$
Then the state after time $\Delta t$ is given by
$$
\mathrm{e}^{-iH\Delta t}\psi = \mathrm{e}^{-i\tilde H_1\Delta t}\cdot...\cdot\mathrm{e}^{-i\tilde H_n\Delta t}\psi_1\otimes...\otimes\psi_n = (\mathrm{e}^{-iH_1\Delta t}\psi_1)\otimes...\otimes(\mathrm{e}^{-iH_n\Delta t}\psi_n)
$$
where the first equality follows from $[\tilde H_i,\tilde H_j] = 0$.
$|\phi(1)\rangle \otimes |\chi(2)\rangle $ is a cumbersome notation to write ket corresponding to $\psi$ function $\phi(\mathbf r_1)\chi(\mathbf r_2)$, where $\mathbf r_i$ refers to coordinates of the $i$-th subsystem. That's why the order of factors in $\otimes$ product does not matter; the resulting ket corresponds to the same $\psi$ function and is thus the same ket.
On the other hand, $|\phi\rangle \otimes |\chi\rangle $ (without labels) is meant to be read according to different convention; here it is commonly understood that the order of factor signifies the sub-system it refers to. So
$|\phi\rangle \otimes |\chi\rangle $ denotes ket corresponding to $\phi(\mathbf r_1)\chi(\mathbf r_2)$ just as $|\phi(1)\rangle \otimes |\chi(2)\rangle $ does, but:
$|\chi\rangle \otimes |\phi\rangle $ denotes ket corresponding to $\chi(\mathbf r_1)\phi(\mathbf r_2)$ which is not the same. This is because different meaning of the $\otimes$ notation is used.
Best Answer
$\lvert A\rangle \langle B \rvert$ is the tensor of a ket and a bra (well, duh). This means it is an element of the tensor product of a Hilbert space $H_1$ (that's where the kets live) and of a dual of a Hilbert space $H_2^\ast$, which is where the bras live. Although for Hilbert spaces their duals are isomorphic to the original space, this distinction should be kept in mind. So we can "feed" a ket $\lvert \psi\rangle$ from $H_2$ to the bra in $\lvert \phi\rangle\otimes \langle\chi\rvert \in H_1\otimes H_2^\ast$, and are left with a state in $H_1$ given by $\langle \chi \vert \psi\rangle \lvert \phi\rangle$. The usual use case for such a tensor product is when $H_1=H_2$ to construct a map from $H_1$ to itself, e.g. the projector onto a state $\lvert \psi \rangle$ is given by $\lvert\psi\rangle \langle \psi \rvert$.
In general, a tensor in $H_2 \otimes H_1^\ast$ corresponds to a linear operator $H_1\to H_2$. In the finite-dimensional case, these are all linear operators, in the infinte-dimensional case, this is no longer true, e.g. $H^\ast \otimes H$ are precisely the Hilbert-Schmidt operators on $H$.
In constract, a tensor $\lvert A\rangle\otimes \lvert B\rangle$ (also just written $\lvert A \rangle \lvert B\rangle$) in $H_1\otimes H_2$, although it corresponds to a bilinear map $H_1\times H_2\to\mathbb{C}$ by definition, is usually not meant to denote an operator, but a state. Given two quantum systems $H_1$ and $H_2$, $H_1\otimes H_2$ is the space of the states of the combined system (as for why, see this question).