Your suspicion is correct, even when $A=B$. Consider the Hilbert space of two qubits and let $^{T_A}$ denote the partial transpose with respect to one of them.
Suppose that
$$
A=B=\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}.
$$
Then
$$
A^{T_A}=B^{T_A}=\begin{pmatrix}
1 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
1 & 0 & 0 & 1
\end{pmatrix}.
$$
and we see that $(AB)^{T_A} = I^{T_A} = I$, but $A^{T_A}B^{T_A} = B^{T_A}A^{T_A} = 2 A^{T_A}$ is a rank one matrix.
I am not aware of a simple identity connecting $(AB)^{T_A}$ and $A^{T_A}B^{T_A}$ or $B^{T_A}A^{T_A}$.
A useful computational aid is offered by tensor notation and in some cases tensor networks. Note that operators such as $A$ and $B$ that act on the Hilbert space of a bipartite system can be thought of as tensors with four indices $A_{ijkl}$ and $B_{ijkl}$. Their usual matrix representation is obtained by grouping pairs of indices for example $A_{(ij)(kl)}$, $B_{(ij)(kl)}$ so that $(ij)$ indexes the rows and $(kl)$ indexes the columns. In this view partial transpose swaps the first indices from each pair $A_{(ij)(kl)}^{T_A} = A_{(kj)(il)}$. Thus
$$
(AB)_{(ij)(mn)} = \sum_{kl} A_{(ij)(kl)}B_{(kl)(mn)}\\
(AB)_{(ij)(mn)}^{T_A} = \left(\sum_{kl} A_{(ij)(kl)}B_{(kl)(mn)}\right)^{T_A} = \sum_{kl} A_{(mj)(kl)}B_{(kl)(in)}\\
(A^{T_A}B^{T_A})_{(ij)(mn)} = \sum_{kl} A^{T_A}_{(ij)(kl)}B^{T_A}_{(kl)(mn)} = \sum_{kl} A_{(kj)(il)}B_{(ml)(kn)}
$$
which shows that changing the order of matrix multiplication and partial transpose affects which elements are multiplied together.
