Suppose a prime $p$ satisfies $p^a \mid m$. Then from the given conditions, there exists an infinite subset $A_1$ of $A$ such that $p$ is coprime to every element in $A_1$.
By the pigeonhole principle, there is an infinite subset $A_2$ of $A_1$, such that $x \equiv a \pmod{mn}$, for each element $x \in A_2$, where $a$ is a positive integer and $p \nmid a$.
Since $(m, n) = 1$, we have $(p^a, \frac{mn}{p^a}) = 1$. By the Chinese Remainder Theorem, we know that
$$\begin{cases} x \equiv a^{-1} \pmod{p^a}, \\ x \equiv 0 \pmod{\frac{mn}{p^a}} \end{cases} \quad ①$$
have infinitely many solutions. Among them we take one as $x$. Next define $B_p$ as the set of the first $x$ elements in $A_2$, and $S_p$ as the sum of all elements in $B_p$. Then we have $S_p \equiv a x \pmod{mn}$. By ① we have
$$S_p \equiv a x \equiv 1 \pmod{p^a}, \quad S_p \equiv 0 \pmod{\frac{mn}{p^a}}.$$

Suppose that $m = p_1^{a_1} \cdots p_k^{a_k}$, and for every $p_i$ ($1 \le i \le k$) select a finite subset $B_i$ of $A$, where $B_i \subset A \setminus (B_1 \cup \cdots \cup B_{i-1})$, such that $S_{p_i}$, the sum of all the elements in $B_i$, satisfies
$$S_{p_i} \equiv 1 \pmod{p_i^{a_i}}, \quad S_{p_i} \equiv 0 \pmod{\frac{mn}{p_i^{a_i}}}. \qquad \textcircled{2}$$
Let $B = \bigcup_{i=1}^k B_i$, whose sum of elements then satisfies $S = \sum_{i=1}^k S_i$. According to ②, we have $S \equiv 1 \pmod{p_i^{a_i}}$ ($1 \le i \le k$), and $S \equiv 0 \pmod{n}$. So $B$ is the required subset, and the proof is complete.

Solution 2:

Divide every element in $A$ by $mn$, and let the remainders which occur infinitely be (in order) $\alpha_1, \alpha_2, \dots, \alpha_k$. We claim that
$$(\alpha_1, \alpha_2, \dots, \alpha_k, m) = 1. \qquad \textcircled{3}$$
Otherwise, assume that $p \mid (\alpha_1, \alpha_2, \dots, \alpha_k, m)$. Then $p \nmid n$, since $(m,n) = 1$. By the given conditions we know that there exist infinitely many elements of $A$ not divisible by $p$. But by the definition of $\alpha_1, \alpha_2, \dots, \alpha_k$, the number of such elements are finite. Then by contradiction, ③ holds. Consequently, there are $x_1, x_2, \dots, x_k, y$, satisfying $\alpha_1 x_1 + \alpha_2 x_2 + \dots + \alpha_k x_k - y m = 1$. Choose a suitable positive integer $r$ such that $r n \equiv 1 \pmod{m}$. Then
$$\alpha_1(r n x_1) + \alpha_2(r n x_2) + \dots + \alpha_k(r n x_k) = r n + r m n y.$$
We select in order $r x_i$ elements from $A$ such that the remainders by $mn$ are $\alpha_i$ ($i = 1, 2, \dots, k$). The set of all these elements is acquired.