Isn't this just trivial since we can choose a two numbers $a$ and $b$ not in $S$ such that they have different last digits, but the sum of the elements in $S$ imply that they have the same last digit. Contradiction.

I suppose in the problem it should say "for all $x\in S$", otherwise the term "remaining elements of $S$ would be a bit strange. Suppose $S\subset A$ has $|S|=405$ and $S$ is good. Let $\sigma:=\sum_{x\in S}x$. For all $x\in S$ we have $x\equiv \sigma-x(10)\iff 2x\equiv \sigma (10)$. Hence $\sigma$ is even, $\sigma=2\tau$ and $x\equiv \tau (5)$ for all $x\in S$. But the largest residue class modulo 5 in $A$ has size 404, contradiction.