Consider the set $D = \{x-y \mid x, y \in A\}$. There are at most $101 \times 100 + 1 = 10101$ elements in $D$. Two sets $A_i$ and $A_j$ have nonempty intersection if and only if $t_i - t_j$ is in $D$. So we need to choose the $100$ elements in such a way that the difference for any two elements is not in $D$.

Now select these elements by induction. Choose one element arbitrarily from $S$. Assume that $k$ elements, $k \le 99$, are already chosen. An element $x$ in $S$ that is already chosen prevents us from selecting any element from the set $x+D$, where $x+D = \{x+y \mid y \in D\}$. Thus after $k$ elements are chosen, at most $10101k$ elements are forbidden. Since $S$ has $1000000$ elements, and $10101 \times 99 = 1000000 - 99$, there is always at least one element left to choose for the next $t_{k+1}$. Therefore, we can choose $100$ such elements $t_1, t_2, \dots, t_{100}$ so that the sets $A_j$ are pairwise disjoint.