Solution 1 (By Richard Stong). For a given positive integer $k$, write $10^k m - n = 2^r 5^s t$, where $\text{gcd}(t, 10) = 1$. For large enough values of $k$, the number of times 2 and 5 divide the left-hand side is at most the number of times they divide $n$, hence by choosing $k$ large we can make $t$ arbitrarily large. Choose $k$ so that $t$ is larger than either $m$ or $n$.
Since $t$ is relatively prime to 10 there is a smallest exponent $b$ for which $t \mid (10^b - 1)$. Thus $b$ is the number of digits in the repeating portion of the decimal expansion for $\frac{1}{t}$. More precisely, if we write $tc = (10^b - 1)$, then the repeating block is the $b$-digit decimal representation of $c$, obtained by prepending extra initial zeros to $c$ as necessary. Since $t$ is larger than $m$ or $n$, the decimal expansions of $\frac{m}{t}$ and $\frac{n}{t}$ will consist of repeated $b$-digit representations of $cm$ and $cn$, respectively. Rewriting the identity in the first line as
$$10^k \left( \frac{m}{t} \right) = 2^r 5^s + \frac{n}{t},$$
we see that the decimal expansion of $\frac{n}{t}$ is obtained from that of $\frac{m}{t}$ by shifting the decimal to the right $k$ places and removing the integer part. Thus the $b$-digit representations of $cm$ and $cn$ are cyclic shifts of one another. In particular, they have the same number of occurrences of each nonzero digit.

Solution 2 (By Zhou Xiaodong). Suppose without loss of generality that $m \ge n$. Note that if the desired holds for the pair $(km, kn)$ for some $k$, then it also holds for $(m, n)$. Write $n = 2^a 5^b l$ for some $l$ relatively prime to 10, and note that it suffices to show the desired statement for the pair $(2^b 5^a m, 2^b 5^a n) = (2^b 5^a m, 10^{a+b} l)$. Further, because $10^{a+b} l$ ends with a string of $a+b$ trailing 0's it suffices to show the desired for the pair $(2^b 5^a m, l)$, where $\text{gcd}(l, 10) = 1$. Thus, it suffices to consider $(m, n)$ with $\text{gcd}(n, 10) = 1$.
For such a pair $(m, n)$, we see that $\text{gcd}(10m - n, 10) = 1$, so we may find some $k$ and some $c$ so that
$c(10m - n) = 10^k - 1$, which after rearranging implies that
$$10cm + 1 = 10^k + cn.$$
In addition, we see that $10^k = 10cm - cn + 1 > 10cn - cn = 9cn > cn$, hence cm has exactly k digits and cn has l digits for l $\le$ k. If $cm = \overline{a_1 \cdots a_k}$ and $cn = \overline{b_2 \cdots b_{k+1}}$ are the decimal expansions of cm and cn (where some of the leading digits of cn may be 0), then the above equality yields the equality of decimal expansions
$$\overline{a_1 \cdots a_k 1} = \overline{1 b_2 \cdots b_{k+1}}.$$
We conclude that $a_1 = b_{k+1} = 1$ and $a_i = b_i$ for $2 \le i \le k$, so each non-zero digit appears among $\{a_i\}$ and $\{b_i\}$ the same number of times, hence appears in cn and cm the same number of times.

Since $t$ is relatively prime to $10$ there is a smallest exponent $b$ for which $t \mid (10^b - 1)$. Thus $b$ is the number of digits in the repeating portion of the decimal expansion for $\frac{1}{t}$. More precisely, if we write $tc = (10^b - 1)$, then the repeating block is the $b$-digit decimal representation of $c$, obtained by prepending extra initial zeros to $c$ as necessary. Since $t$ is larger than $m$ or $n$, the decimal expansions of $\frac{m}{t}$ and $\frac{n}{t}$ will consist of repeated $b$-digit representations of $cm$ and $cn$, respectively. Rewriting the identity in the first line as
$$10^k \left(\frac{m}{t}\right) = 2^r 5^s + \frac{n}{t},$$
we see that the decimal expansion of $\frac{n}{t}$ is obtained from that of $\frac{m}{t}$ by shifting the decimal to the right $k$ places and removing the integer part. Thus the $b$-digit representations of $cm$ and $cn$ are cyclic shifts of one another. In particular, they have the same number of occurrences of each nonzero digit.

Suppose without loss of generality that $m \ge n$. Note that if the desired holds for the pair $(km, kn)$ for some $k$, then it also holds for $(m, n)$. Write $n = 2^a 5^b l$ for some $l$ relatively prime to $10$, and note that it suffices to show the desired statement for the pair $(2^b 5^a m, 2^b 5^a n) = (2^b 5^a m, 10^{a+b} l)$. Further, because $10^{a+b} l$ ends with a string of $a+b$ trailing 0's it suffices to show the desired for the pair $(2^b 5^a m, l)$, where $\text{gcd}(l, 10) = 1$. Thus, it suffices to consider $(m, n)$ with $\text{gcd}(n, 10) = 1$.
For such a pair $(m, n)$, we see that $\text{gcd}(10m - n, 10) = 1$, so we may find some $k$ and some $c$ so that
$c(10m - n) = 10^k - 1$, which after rearranging implies that
$$10cm + 1 = 10^k + cn.$$
In addition, we see that $10^k = 10cm - cn + 1 > 10cn - cn = 9cn > cn$, hence $cm$ has exactly $k$ digits and $cn$ has $l$ digits for $l \le k$. If $cm = \overline{a_1 \cdots a_k}$ and $cn = \overline{b_2 \cdots b_{k+1}}$ are the decimal expansions of $cm$ and $cn$ (where some of the leading digits of $cn$ may be $0$), then the above equality yields the equality of decimal expansions
$$\overline{a_1 \cdots a_k} \overline{1} = \overline{1} \overline{b_2 \cdots b_{k+1}}.$$
We conclude that $a_1 = b_{k+1} = 1$ and $a_i = b_i$ for $2 \le i \le k$, so each non-zero digit appears among $\{a_i\}$ and $\{b_i\}$ the same number of times, hence appears in $cn$ and $cm$ the same number of times.