Digit products and sums

Write $m = 100N + 10d_{10} + d_1$ where $N = \left\lfloor\frac{m}{100}\right\rfloor$ and $0 \leq d_{10}, d_1 \leq 9$ are the tens and ones digits, respectively, of $m.$ The rest of the proof will follow by several smaller arguments, so I'll number them to keep them separate and organized.

1. We note that $P(n),$ as the product of digits, for any natural number $n,$ is either $0$ or a product of primes less than $10,$ so that we have $11 \text{ divides } P(n) \iff P(n) = 0.$ Since $11$ divides $990 = P(m) - P(m+2),$ $P(m) = 0 \iff 11 \text{ divides } P(m) \iff 11 \text{ divides } P(m+2) \iff P(m+2)=0 \iff 990 = P(m)-P(m+2) = 0$ Obviously, this is a contradiction, so we can conclude that $P(m),P(m+2) \neq 0,$ so none of the digits of either $m,m+2$ is zero, and in particular, $d_1 \neq 0, 8.$
2. If $d_1 \leq 7$ , then $d_1 + 2 \leq 9$ , so $P(m) = d_1P(10N+d_{10}) \leq (d_1+2)P(10N+d_{10}) = P(m+2) \implies 990 = P(m)-P(m+2) \leq 0$ which is once again a contradiction, so we conclude that $d_1 > 7.$ Combining this with the result of Step 1, we conclude $d_1 = 9.$
3. Then $m+2 = 100N + 10d_{10} + 9 + 2 = 100N + 10(d_{10} + 1) + 1$ and by the result of step 1 again, $d_{10} + 1 \neq 10,$ so in particular $d_{10} + 1 \leq 9$ is the tens digit of $m+2.$ This allows us to write $990 = P(m) - P(m+2) = P(N)(d_{10}\cdot 9) - P(N)((d_{10}+1)\cdot 1) = P(N)(8d_{10} - 1),$ so $P(N)(8d_{10} - 1) = 990 \equiv 0 \pmod{11} \implies N=0 \text{ or } d_{10} \equiv 7 \pmod{11} \implies N=0 \text{ or } d_{10} = 7.$ Now, if $N=0$ then $990 \leq P(m) = d_{10}\cdot 9 \leq 9\cdot 9 = 81$ which is another contradiction, so we can conclude $d_{10} = 7,$ and putting this back into a previous equation in this step, $990 = P(N)(8(7)-1) \implies P(N) = 18.$ Since $18 = 9\times 2 = 6\times 3 = 3\times 3 \times 2,$ the digits of $N$ must be any number of $1\text{s}$ along with one of the following cases: one $9$ and one $2;$ one $6$ and one $3;$ or two $3\text{s}$ and one $2.$
4. Since $m = 100N + 79 \equiv 0 \pmod{99}$ we can solve for $N \equiv 20 \pmod{99}.$ Here, it will be useful to list a few more residues, so $N \equiv 20, 119, 218 \pmod{99}$
5. If we suppose $N$ has $M$ digits and use $10^2 \equiv 1 \pmod{99}$ to compute $10^{M-1} + 10^{M-2} + \cdots + 10 + 1 \equiv 10\left\lfloor\frac{M}{2}\right\rfloor + \left\lfloor\frac{M+1}{2}\right\rfloor \pmod{99}$ Then we can compute $N \pmod{99}$ by listing this result for increasing values of $M$ and adding $(d-1)$ or $10(d-1)$ for each set of the possible digits of $N$ not equal to $1.$ This is a little bit of a slog (though it can be made a little more efficient, I didn't find a good reason to make such an argument), but we end up finding the smallest value of $N$ is given by $M=13$ using the digits $6$ and $3,$ which works because $10\left\lfloor\frac{13}{2}\right\rfloor + \left\lfloor\frac{13+1}{2}\right\rfloor + 10(6-1) + (3-1) = 119$ is one of the residues for $N \pmod{99}$ we found earlier. It doesn't matter for the final answer, but for those who are interested, this gives $N = 1111111111163$ so that the smallest value of $m$ is $m = 111111111116379$

Finally, for this value of $m,$ we have $S(m) = 11\cdot 1 + 6 + 3 + 7 + 9 = 36 \\ S(m+2) = 11\cdot 1 + 6 + 3 + 8 + 1 = 29$ giving $S(m) + S(m+2) = 36 + 29 = \boxed{65}$