Sketch. $S$ has $10$ atoms, because otherwise we could add some $100$-digit number to $S$ and increase the number of elements in $S$. Let $A = \{a_1,\ldots,a_{10}\} \subseteq S$ be the set of $10$ atoms. By induction it can be shown that every element $s \in S$ can be expressed as a linear combination $s = \lambda_1a_1+\lambda_2a_2+\cdots +\lambda_{10}a_{10}$, for integers $0\le \lambda_i \le 9$ and $1 \le \lambda_1+ \cdots +\lambda_{10} \le 9$. Furthermore, if $S$ is maximal then every such linear combination must be an element of $S$. So enumerating all the possible vectors $(\lambda_1,\lambda_2,\ldots,\lambda_{10})$ by a ball-and-urn argument gives an upper bound of $\binom{9+10}{9} - 1$. This can be achieved by defining $a_i = 10^{99} + 2^{i-1}$ for $i= 1,2,\ldots,10$, and setting $S$ equal to all linear combinations of the $a_i$.

The only thing that has to be changed is the example you gave, as $10^{99}+2^{i-1}$ doesn't give distinct linear combinations (simply try $2\cdot 2^{i-1}=2^i$). But $10^{99}+10^i$ constitutes a good one, because of the uniqueness of base $10$ representation.