The maximum size is $9$, e.g. take $S=\{12,13,14,15,16,26,36,46,56\}=\{10a+b\in T|\{a,b\}\cap\{1,6\}\ne\emptyset\}$. The problem can be formulated as follows: Find a maximal subset of $E(K_6)$ that covers all vertices and contains no 1-factor. To see that $9$ is best, we first prove $|S|\le 10$. Take an 1-factorisation of $K_6$. Then $S$ can only contain at most two edges of each of the five 1-factors, so $|S|\le 10$. Now it is easy to see that a 10-element set $S$ without an 1-factor must consist of all the edges of a $K_5$, so it wouldn't cover all vertices (=digits) as requested.