The examples listed in the question are not the only ones -- for instance, if $k=l=5$ then you can take $S=\{0,1,2,4,8\}$.
I note that there are a few ways of constructing more solutions from a given solution. For instance, if $S$ is a solution in $\mathbb{Z}/2l\mathbb{Z}$ then $t+nS$ is a solution in $\mathbb{Z}/2ln\mathbb{Z}$ for any $t$ and $n$ such that either $n$ is odd or all elements of $S$ are even. Thus, the solutions listed in the question all come from the solution $\{0,2,4,6,...,2k−2\}$ with $k=l$.
Next, if $S_1$ and $S_2$ are two solutions in $\mathbb{Z}/2l\mathbb{Z}$ with $l$ even such that all elements of $S_1$ are even and all elements of $S_2$ are odd, then $S_1\cup S_2$ is a solution. Thus, for example, from the solution $\{0,2,4\}$ with $l=3$ we can build the solution $\{0,1,4,5,8,9\}$ with $l=6$; and from $\{0,2,4\}$ with $l=3$ and $\{0,2,4,6,8\}$ with $l=5$ we can build the solution $\{1,21,41\}\cup\{0,12,24,36,48\}=\{0,1,12,21,24,36,41,48\}$ with $l=30$.
There are probably further ways to build new solutions from old ones, and it seems that one should take account of these in order to have any hope of describing all solutions.