$\newcommand\N{\mathbb N}$
Define the nondecreasing sequences $(A_n)_{n\in\N}$ and $(B_n)_{n\in\N}$ of subsets of $\N=\{1,2,\dots\}$ recursively as follows:
$$A_1:=\{1\},\quad B_1:=\{4\};$$
for $n\ge2$,
$$
(A_n,B_n):=\left\{
\begin{aligned}
(A_{n-1},B_{n-1})&\text{ if }c_n\in A_{n-1}B_{n-1},\\
(A_{n-1},B_{n-1}\cup\{c_n\})&\text{ if }c_n\notin \N B_{n-1},\\
(A_{n-1}\cup\{a^*_n\},B_{n-1})&\text{ if }c_n\in\N B_{n-1}\setminus A_{n-1}B_{n-1}\\
&\text{ and }a^*_n>\max A_{n-1}, \\
(A_{n-1},B_{n-1}\cup\{c_n\})&\text{ if }c_n\in\N B_{n-1}\setminus A_{n-1}B_{n-1}\\
&\text{ and }a^*_n\le\max A_{n-1},
\end{aligned}
\right.
$$
where
$$a^*_n:=\min(\N\cap(c_n/B_{n-1})).$$

Let now
$$A:=\{a_1,a_2,\dots\}:=\bigcup_{n\in\N}A_n,\quad B:=\{b_1,b_2,\dots\}:=\bigcup_{n\in\N}B_n,$$
where $a_1<a_2<\cdots$ and $b_1<b_2<\cdots$.

Then the product of $A$ and $B$ equals $C:=\{c_1,c_2,\dots\}$, where $A,B,C$ are considered multisets. That is, for each $c\in C$ there is a unique pair $(a,b)\in A\times B$ such that $c=ab$.

The identity in question now follows.

For an illustration, note that, in particular,
$$A_{50}=\{1, 2, 4, 8, 16\},$$
$$B_{50}=\{4, 6, 9, 10, 14, 15, 21, 22, 25, 26, 27, 33, 34, 35, 38, 39, 45, 46, \
49, 51, 55, 57, 58, 62, 63, 65, 69\}.$$