We will show that the only solution is a function $f$ such that
$$f(m) = \begin{cases} p, & \text{if } m \text{ is a non-trivial power of a prime } p, \text{ i.e. } m = p^k, k \ge 1, \\ 1, & \text{otherwise.} \end{cases}$$
Number $1$ has the unique divisor $1$, hence plugging $m = 1$ in the given equality we get $f(1) = 1$.
Let $m = p$ be a prime. Then
$$f(1) \cdot f(p) = p \quad \text{i.e.} \quad f(p) = p.$$
For $m = p^2$ we get
$$f(1) \cdot f(p) \cdot f(p^2) = p^2 \quad \text{i.e.} \quad f(p^2) = p$$
and in general for $k > 1$ and $m = p^k$
$$f(1) \cdot f(p) \cdot f(p^2) \cdots f(p^k) = p^k.$$
By induction, $f(p^k) = p$ for all positive integers $k$.
Now let us consider positive integer $m$ with at least two distinct prime factors whose factorization is $m = p_1^{\alpha_1} p_2^{\alpha_2} \dots p_k^{\alpha_k}$, where $k \ge 2$ and $\alpha_i \ge 1$ for every $i \in \{1, 2, \dots, k\}$. The divisors of $m$ include the powers
$$p_1, p_1^2, \dots, p_1^{\alpha_1}, p_2, p_2^2, \dots, p_2^{\alpha_2}, \dots, p_k, p_k^2, \dots, p_k^{\alpha_k},$$
of its prime factors but the product of the corresponding function values
$$\begin{aligned} & f(p_1)f(p_1^2)\dots f(p_1^{\alpha_1})f(p_2)f(p_2^2)\dots f(p_2^{\alpha_2})\dots f(p_k)f(p_k^2)\dots f(p_k^{\alpha_k}) \\ &= \underbrace{p_1p_1\dots p_1}_{\alpha_1} \underbrace{p_2p_2\dots p_2}_{\alpha_2} \dots \underbrace{p_kp_k\dots p_k}_{\alpha_k} = p_1^{\alpha_1}p_2^{\alpha_2}\dots p_k^{\alpha_k} = m \end{aligned}$$
yields $m$. This means that the product of all the other function values (none of which is a non-trivial power of a prime), including $f(m)$, is equal to $1$ so all the other function values, including $f(m)$, are equal to $1$.
This specifies $f$ uniquely and also shows that it has the desired properties.