OK, by Seva's request I'm getting somewhat more serious :) Fix $a$, $b$. Take large $M$ to be chosen shortly. Take a $3MN$ digit number $n$ (something from $0$ to $10^{3MN}-1$, written with head zeroes if necessary) and split its decimal representation into pieces of length $3M$ where $M$ is large but fixed. Let us call these $3M$-digit pieces $n_k$ so $n=\sum_{k=0}^{N-1}10^{3Mk}n_k$. Consider $n_k$ as independent random variables uniformly distributed in the set $\{0,1,2,\dots,10^{3M}-1\}$. By the law of large numbers, with probability close to $1$, there are about $N\cdot 10^{-2M}$ pieces $n_k$ that start and end with $M$ zeroes (with the middle $M$ digits being anything). Call the numbers $n$ that satisfy this property typical. The typical numbers can be split into groups according to exactly which $k$ correspond to such pieces (denote by $K$ the set of all such $k$) and what number is formed by the digits outside these groups (that number is $n'=\sum_{k\notin K}10^{3Mk}n_k$). We need to show that in each group $G=G(K,n')$ of typical numbers the portion of the solutions is small.

As Fedor explained, if $M$ exceeds the number of digits in $a$, we have $s(an)=s(an')+\sum_{k\in K}s(an_k)$ and similarly for $b$. Thus
$$
s(an)-s(bn)=s(an')-s(bn')+\sum_{k\in K}X_k
$$
with
$$
X_k=s(an_k)-s(bn_k)
$$
The random variables $X_k$ are i.i.d. and their distribution is completely determined by $a,b$, and $M$.

Suppose that the equality $s(an)=s(bn)$ fails for at least one $n\ge 0$. Then (since it holds for $n=0$) $X_k$ is not a constant for sufficiently large $M$ and we can use the following

*Probabilistic Claim.* If $X$ is any fixed integer-valued random variable with finitely many values that is not constant, then $\lim_{Q\to\infty}\sup_{S\in\mathbb Z} P\{\sum_{k=1}^Q X_k+S=0\}=0$ where $X_k$ are i.i.d. random variables equidistributed with $X$.

Since the cardinality $Q$ of $K$ for every typical group is huge when $N$ is large enough, we conclude that the above displayed equation has very little chance to hold in every
typical group $G(K,n')$ and, thereby, overall.

However, if $s(an)=s(bn)$ for all $n$, then $a$ and $b$ differ only by the number of zeroes in the end.

I apologize if this edit rendered some comments meaningless.