By "without trivial commutativities" I mean "$bc\neq cb$ for any $b,c \in S \setminus \{a\}$, where $a$ is the identity and $b\neq c$.
I found out how to calculate the element $x_{ij}$ of the $n\times n$ Cayley table for the general* symmetric case: $x_{ij} = mod(i+j,n)$. However, I am clueless about how to calculate it for the asymmetric case (manually permuted here to achieve minimal commutativity).
Probably, defining the general* asymmetric case itself is part of the solution. The result of the formula above is given in the Figure for each pair of elements, and the correct elements are marked as ok in the last table.
general* -> general for any finite number of elements that is suitable to define such a group
The only groups that satisfy the condition $bc \ne bc$ for all $b,c \ne 1$ and $b \ne c$ are the groups of orders $1$ and $2$, which are abelian and satisfy the condition trivially.
Otherwise, if there exists and element with $g \ne g^{-1}$ then the condition fails with $b=g$ and $c=g^{-1}$. If there is no such element, then by a well-known result $G$ is abelian.
A more interesting condition would be $bc = cb$ if and only if $b$ is a power of $c$ or $c$ is a power of $b$. That holds in some examples, such as dihedral groups of twice odd order, and also in some infinite groups, such as free groups.