My point was that you can prove that S is not zero; there is no need to assert it based on an assumption about unstated rules. I personally enjoy puzzles and believe that they are good tools for teaching how to solve problems, but making unsupported assumptions defeats the purpose.
Although I did not find it the obvious way to start, you certainly may start by letting x = the number represented by STONE, y = the number represented by DISEASED, z = the number represented by SEE, and p = the number represented by SONUAO.
[MATH]\dfrac{y}{x} = z + \dfrac{p}{x} \implies y = xz + p. [/MATH]
[MATH]S \ne 0 \implies 0 < S. [/MATH] I shall let the OP prove this.
[MATH]\therefore x > 10,000S,\ z \ge 100S,\text { and } p > 100,000 \implies xz + p > 1,000,000S^2 + 100,000.[/MATH]
By the rules of the puzzle, D < S or D > S.
[MATH]D < S \implies y < 1,000,000(S - 1) + 999,999 = 1,000,000S - 1.[/MATH]
[MATH]1 \le S \implies S \le S^2 \implies 1,000,000S \le 1,000,000S^2 \implies[/MATH]
[MATH]y < 1,000,000S - 1 < 1,000,000S \le 1,000,000S^2 < 1,000,000S^2 + 100,000 < xz + p.[/MATH]
But that is a contradiction. So D > S.
Summarizing
[MATH]1 \le S < D \le 9 \implies 1 \le S \le 8, \ 2 \le D \le 9, \text { and } 1 \le D - S \le 8.[/MATH]
So the OP is correct, but for the wrong reason.