There are 348364800 such functions.
We first claim that condition (iii) can be replaced by the completely multiplicative condition. This means
$$f(p_1^{a_1} p_2^{a_2} \cdots p_s^{a_s}) = f(p_1)^{a_1} f(p_2)^{a_2} \cdots f(p_s)^{a_s}$$
for any distinct primes $p_1, p_2, \dots, p_s$ and positive integers $a_1, a_2, \dots, a_s$. We prove this by induction on $a_1 + a_2 + \cdots + a_s$.
The base cases $a_1 + a_2 + \cdots + a_s = 0, 1$ hold trivially. Assume the result holds whenever the sum of exponents is less than $k$. Consider $n = p_1^{a_1} p_2^{a_2} \cdots p_s^{a_s}$ where $a_1 + a_2 + \cdots + a_s = k$. Note that $n$ is composite, and so $g(n), h(n) > 1$. Since $g(n)h(n) = n$, the sum of exponents in the prime factorizations of $g(n)$ and $h(n)$ are both smaller than $k$. Thus, by the inductive hypothesis, $f(g(n))$ and $f(h(n))$ can be decomposed into a product of $f(p_j)$'s. Clearly, this proves
$$f(n) = f(p_1^{a_1} p_2^{a_2} \cdots p_s^{a_s}) = f(p_1)^{a_1} f(p_2)^{a_2} \cdots f(p_s)^{a_s}.$$

By induction, we have shown that $f$ is completely multiplicative. Conversely, if $f$ is completely multiplicative, condition (iii) holds trivially.
Let $\mathcal{P}$ be the set of all primes from 1 to 100, and let $\mathcal{C} = S - (\mathcal{P} \cup \{1\})$ be the set of all composites. For any $n \in \mathcal{C}$, since $g(n), h(n) > 1$, we must have $f(g(n)), f(h(n)) > 1$. Therefore, $f(n) = f(g(n))f(h(n)) \in \mathcal{C}$. This implies $f(\mathcal{C}) = \mathcal{C}$. As $f(1) = 1$, by the bijective condition, we must have $f(\mathcal{P}) = \mathcal{P}$. Thus, it remains to assign distinct prime values to $f(p)$ for $p \in \mathcal{P}$. Afterwards we can easily compute $f(n)$ by the completely multiplicative condition, while the bijective condition is always satisfied.
Now, the only constraint is that $f(n) \in S$ for any $n \in S$. For example, since
$$100 \ge f(64) = f(2)^6$$
and $f(2) \ne 1$, we must have $f(2) = 2$. Similarly, by considering $f(81)$, we can prove that $f(3) = 3$. Next, since
$$100 \ge f(75) = f(3)f(5)^2,$$
we have $f(5) \le 5$, and hence $f(5) = 5$. Similarly, by considering $f(98), f(99)$ and $f(91)$, we can prove that $f(p) = p$ for $p = 7, 11, 13$.
Next, by considering $f(85)$ and $f(95)$, we can only prove that $f(17), f(19) \le 20$. Therefore, we have $\{f(17), f(19)\} = \{17, 19\}$. There is no further constraint on $f(17)$ and $f(19)$ since we can verify that $f(n) \in S$ for any $n$ being multiples of 17 or 19, as long as $f(17), f(19) \le 20$.
Similarly, by considering $f(92)$, we find that $f(23) = 23$.
By considering $f(87)$ and $f(93)$, we find that $\{f(29), f(31)\} = \{29, 31\}$.
By considering $f(74), f(82), f(86)$ and $f(94)$, we find that $37, 41, 43, 47$ can be mapped to any permutation of themselves.
Lastly, the images of $53, 59, 61, 67, 71, 73, 79, 83, 89, 97$ can be any permutation of themselves.
In view of the above arguments, we can choose the images within each group in an arbitrary way. Each choice corresponds to a unique function satisfying all requirements. Therefore, the number of such functions is
$$$$
2! $\times$ 2! $\times$ 4! $\times$ 10! = 348364800.