Given a bipartite state $\rho_{AB}$ living in the Hilbert space $\mathcal H(A\otimes B)$ we can always define two local states on $A$ and $B$ respectively by taking the appropriate partial traces: $$\rho_A=\mathrm{tr}_B[\rho_{AB}], \quad \rho_B=\mathrm{tr}_A[\rho_{AB}]. $$ If the state $\rho_{AB}$ is separable, then $\rho_{AB}=\rho_A\otimes \rho_B$. In the more general case of an entangled state, since taking the partial trace corresponds to discarding part of the system any correlations between such a subsystem and its complement are lost, so $\rho_{AB}\ne \rho_A\otimes \rho_B$ and more generally there can be no pair of local states $\sigma_A,\tau_B$ such that $\rho_{AB}=\sigma_A\otimes \tau_B$.
But what about states that have classical correlations? I give an example below.
Example: Given an ensemble $\mathcal E=\{p_j,\rho_Q(j)\}$ consider the state $$\rho_{QX}=\sum_j p_j\rho_Q(j)\otimes |j\rangle\langle j|_X $$ where $\{|j\rangle_X\}$ is an orthonormal basis of $\mathcal H(Y)$. This state is correlated as the index $j$ connects the states on $Q$ and on $X$, but it is not entangled since it can be written as $$\rho_{QX}=\sum_j p_j\rho_Q(j)\otimes \rho_X(j) . $$
The partial traces give $$\rho_Q=\sum_j p_j\rho_Q(j), \quad \rho_X=\sum_j p_j|j\rangle\langle j|_X $$ and so $\rho_Q\otimes \rho_X=\sum_j p_j^2\rho_Q(j)\otimes |j\rangle\langle j|_X$. This looks almost like $\rho_{QX}$ apart from that coefficient $p_j^2$ that seems to reflect the fact that when we separate $\rho_{QX}$ we can choose to 'put' $p_j$ either in local state on $Q$ or on the on $X$.