One option is as follows: You write both numbers in base $6$, if necessary writing the smaller one with leading zeroes so that the length matches the larger one. Then you map each digit to one of the permutations in $S_3$, mapping the digit $0$ to the identity permutation (how you map the others is not relevant, as long as you stay consistent). Then you multiply the permutations for corresponding digits of the two numbers, and turn the result back into a digit. Finally you interpret the result again as base $6$ representation of an integer.
For example, assume the following mapping:
\begin{align}
0 &\to \operatorname{id}\\
1 &\to (12)\\
2 &\to (13)\\
3 &\to (23)\\
4 &\to (123)\\
5 &\to (132)
\end{align}
Now consider for example $16*2$. Rewriting in base 6 gives $24_6*02_6$. Now you go digit-wise:
\begin{align}
2*0 &\to (13)\operatorname{id} = (13) \to 2\\
4*2 &\to (123)(13) = (23) \to 3
\end{align}
Finally we find that $23_6 = 15$. Therefore $16*2=15$
This is a group where the neutral element is $0$, and the inverse element is obtained by replacing each $4$ by $5$ and vice versa in the base $6$ representation. It is also non-commutative, as seen by the fact that $2*16 = 13 \ne 16*2$.