The first trick here is to think of the multiplication table as the transition table of an automaton $A$ with each state representing a letter in your multiplication table, but not worrying about acceptance yet. So the letters on the left and in the body of the table are actually states -- it would be more accurate to write them as $q_a, q_b, q_c$, but I won't. The letters across the top are inputs.
Then construct the automaton $A_T$ ("$T$" for transpose) for reverse multiplication by transposing $A$:
$\qquad \displaystyle\begin{array}{c|ccc}
A_T & a & b & c \\
\hline
a & a & c & b \\
b & a & a & c \\
c & c & b & a
\end{array}$
So $A(abc)$ takes you to state $c$, and likewise $A_T(cba)$ moves into state $a$ of $A_T$, as you note.
However, $A_T$ assumes you are going right-to-left, and we still want to go left-to-right. So the second trick is to reverse the automaton (not the multiplication, which would just get us back were we started), by reversing all the arrows, which leads to a non-deterministic automaton $A_{TR}$ given by the transition table below, with subsets indicated by concatenated letters to keep the chicken scratching down, so $ac$ is really $\{a,c\}$. (hope I got it all right -- seems to work).
$\qquad \displaystyle\begin{array}{c|ccc}
A_{TR} & a & b & c \\
\hline
a & ab & b & c \\
b & ∅ & c & a \\
c & c & a & b \\
\hline
ab & ab & bc & ac \\
bc & c & ac & abc \\
ac & abc & ab & bc \\
abc & abc & abc & abc \\
∅ & ∅ & ∅ & ∅
\end{array}$
You can interpret this as a non-deterministic automaton with only the three rows above the line or a determinized version with all 8 rows.
Finally, the machine to solve the problem is the cross-product automaton of the original $A$ and $A_{TR}$, that is $A \times A_{TR}$ to perform the intersection behavior of the two automata (we don't need $A_T$ any more). $A \times A_{TR}$ has states that are pairs like $\langle a,ac\rangle$. The transition function runs $A$ and $A_{TR}$ independently. A single start state $\langle 1,1 \rangle$ goes into $\langle a,a \rangle$ under input $a$, into $\langle b,b \rangle$ under input $b$, etc.
Accepting states in the non-deterministic version are $\langle a,a \rangle$ etc. In the deterministic version, accepting states are pairs in which the first component is $\in$ of the second component set, such as $\langle a,a \rangle$ or $\langle b,bc \rangle$.
$A \times A_{TR}$ augmented and determinized as shown has $25=3\cdot 8+1$ states, so forgive me if I don't write it out in detail. But the non-deterministic version has only $10=3\cdot 3+1$ states.