Explicitly write out what the trace is supposed to be:
$$
\text{Tr}(S_{AB}S_{BC}S_{CD}\rho_a\otimes\rho_a\otimes\rho_b\otimes\rho_b)=\sum_{i,j,k,l}\langle ijkl|S_{AB}S_{BC}S_{CD}\rho_a\otimes\rho_a\otimes\rho_b\otimes\rho_b|ijkl\rangle.
$$
Now you can apply the swaps to the basis elements
$$
=\sum_{i,j,k,l}\langle jikl|S_{BC}S_{CD}\rho_a\otimes\rho_a\otimes\rho_b\otimes\rho_b|ijkl\rangle
$$
and keep going...
$$
=\sum_{i,j,k,l}\langle jkli|\rho_a\otimes\rho_a\otimes\rho_b\otimes\rho_b|ijkl\rangle
$$
So, this is the same thing as
$$
\sum_{ijkl}\langle j|\rho_a|i\rangle\langle i|\rho_b|l\rangle\langle l|\rho_b|k\rangle\langle k|\rho_a|j\rangle.
$$
It's worth a comment about how I chose the ordering of those terms. I just started with the first index ($j$). Because it's closing index was $i$, the next term I wrote down started with $i$, and so on. Now, knowing the result I want, I'll use the commutativity of multiplication to choose a slightly different order:
$$
\sum_{ijkl}\langle k|\rho_a|j\rangle\langle j|\rho_a|i\rangle\langle i|\rho_b|l\rangle\langle l|\rho_b|k\rangle.
$$
Next, notice that there's lots of completeness relations appearing here: $\sum_j|j\rangle\langle j|=I$. Thus, this is the same as
$$
\sum_k\langle k|\rho_aI\rho_aI\rho_bI\rho_B|k\rangle=\text{Tr}(\rho_a^2\rho_b^2).
$$
The other one will work just the same.