Instead of sequence $a_n$, we'll use notation with the function $f(n)$ with same properties.

There exists only one such function: $f(n) = n$. We'll solve the problem with many separate facts.

Fact 1: $f(1) = 1$

Proof. According to a) it holds $f(1) = f(1)f(1) = f(1)^2$. Since $f(1)$ is positive integer, it can't be $f(1) = 0$, so it must be $f(1) = 1$.

Fact 2: Function $f$ is bijective.

Proof. Firstly, we'll show that $f$ is injective. Let $a \neq b$ be two arbitrary positive integers and let's assume $f(a) = f(b)$. Since $\{1, 2, ..., n\} = \{f(1), f(2), ..., f(n)\}$ holds for infinitely any positive integers $n$, it holds for some integer greater than $a$ and $b$. Then, since $f(a) = f(b)$, set $\{f(1), f(2), ..., f(n)\}$ contains $n-1$ or less (different) elements, but according to b), it contains $n$ elements.

Secondly, we'll show that $f$ is surjective. Let $c$ be arbitrary integer and let's assume that $f(n) \neq c$ for all positive integers $n$. Similarly as in first part of proof, let's take positive integer $n$ such that $\{1, 2, ..., n\} = \{f(1), f(2), ..., f(n)\}$ holds. Since $c \in \{1, 2, ..., n\}$, $c$ is also element of the set $\{f(1), f(2), ..., f(n)\}$, so there exists positive integer $m \le n$ such that $f(m) = c$.

Fact 3: Positive integer $n$ is prime if and only if $f(n)$ is prime.

Proof. Let's assume that $n$ is prime, but $f(n)$ isn't. Then it must be $f(n) = a'b' = f(a)f(b) = f(ab)$, where $a', b'$ are positive integers greater 1, and $a, b$ are unique positive integers such that $f(a) = a'$, $f(b) = b'$ (they exist since $f$ is bijective). Since $f$ is injective, $f(1) = 1$ and $a', b'$ are not equal to 1, integers $a, b$ are also not equal to 1. Since $f$ is injective and $f(n) = f(ab)$, we have $n = ab$, so $n$ is composite.

Let's assume that $f(n)$ is prime, but $n$ isn't. Then there exist positive integers $a, b$ greater than one such that $n = ab$. From there we have $f(n) = f(ab) = f(a)f(b)$. Again from injectivity of $f$ and $f(1) = 1$, we see that $f(n)$ is product of two integers greater than 1.

Fact 4: If $n = p_1^{\alpha_1} p_2^{\alpha_2} ... p_k^{\alpha_k}$ is unique factorization of positive integer $n$, then
$$f(n) = f(p_1)^{\alpha_1} f(p_2)^{\alpha_2} ... f(p_k)^{\alpha_k}$$
is unique factorization of positive integer $f(n)$.

Proof. From multiple use of the condition a) we get identity $f(n) = f(p_1)^{\alpha_1} f(p_2)^{\alpha_2} ... f(p_k)^{\alpha_k}$. From fact 3, numbers $f(p_i)$ are prime. Since $f$ is injective, none of two numbers $f(p_i)$ and $f(p_j)$ are equal.

Fact 5: (Technical result) For all positive integers $y < x$ there exist positive integer $n_o$ such that for all positive integers $n > n_o$ holds inequality
$$y^{n+1} < x^n.$$

Proof. It is sufficient to prove the fact only for consecutive integers $y$ and $y+1$ (because we'll have $y+1 < (y+1)^n \le x^n$. By binomial theorem we have
$$(y+1)^n \ge y^n + n y^{n-1} = y^{n-1}(y+n).$$
Thus if we define $n_o = y^2 - y + 1$, then for all $n \ge n_o$ we have
$$(y+1)^n \ge y^{n-1}(y+n) \ge y^{n-1}(y+n_o) = y^{n-1}(y^2+1) > y^{n+1}.$$
*Another proof.* Inequality is equivalent to
$$\left(\frac{y}{x}\right)^n > y.$$
The fact follows from the fact that the expression on the left hand side is increasing and it is unbounded, while the right hand side is fixed.

Fact 6: For all prime numbers $p$ we have $f(p) \le p$.

Proof. Let $p_1, p_2, \dots, p_n, \dots$ be the increasing sequence $2, 3, 5, 7, \dots$ of all primes. Let's take arbitrary prime number $p_n$. From the Fact 3 we have that $f(p_n)$ is also prime. Let's take positive integer $n_0$ as the integer from the Fact 5, for positive integers $y = p_n < p_{n+1} = x$. Since b) holds for infinitely many positive integers, it holds for some positive integer $N$ such that $\{1, 2, \dots, N\} = \{f(1), f(2), \dots, f(N)\}$, and such that $N \ge p_n^{n_0}$. Let $\alpha$ be the greatest positive integer such that $p_n^{\alpha} \le N$. From definitions of $N$ and $\alpha$ we have $\alpha \ge n_0$.

In set $\{1, 2, \dots, N\}$ we'll observe all positive integers which are $\alpha^{th}$ power of a prime number. Since $N \ge p_n^{\alpha}$, we have that $p_n^{\alpha}$ is in that set. It is easy to see that all numbers $p_1^{\alpha}, p_2^{\alpha}, \dots, p_{n-1}^{\alpha}$ are also in that set. On the contrary, number $p_{n+1}^{\alpha}$ is not in that set, because from the definition of $\alpha$ and $N$ respectively we have $N < p_n^{\alpha+1} \le p_{n+1}^{\alpha}$ (remember Fact 5 and $\alpha \ge n_0$). Similarly, neither of the numbers $p_m^{\alpha}$ (for $m > n$) is not in the set $\{1, 2, \dots, N\}$.

Let us now observe all positive integers which are $\alpha^{th}$ power of a prime and they are in the set $\{f(1), f(2), \dots, f(N)\}$. According to Fact 4, we have that $f(n)$ is $\alpha^{th}$ power of a prime. From that and from previous paragraph we conclude that only such numbers are $f(p_1^{\alpha}), f(p_2^{\alpha}), \dots, f(p_n^{\alpha})$.

Now we have $\{p_1^{\alpha}, p_2^{\alpha}, \dots, p_n^{\alpha}\} = \{f(p_1^{\alpha}), f(p_2^{\alpha}), \dots, f(p_n^{\alpha})\}$. Thus $f(p_n^{\alpha}) \in \{p_1^{\alpha}, p_2^{\alpha}, \dots, p_n^{\alpha}\}$, so $f(p_n^{\alpha}) = p_i^{\alpha}$ for some $1 \le i \le n$, which implies $f(p_n^{\alpha}) = p_i^{\alpha}$ for some $1 \le i \le n \Rightarrow f(p_n) = p_i \le p_n$, which completes the proof.

Fact 7: For every positive integer we have $f(n) = n$.

Proof. From Fact 3 we have $f(p)$ if and only if $p$ is prime. Let $p_1, p_2, \dots, p_n, \dots$ be the increasing sequence $2, 3, 5, 7, \dots$ of all prime numbers. From fact 6 we have $f(p_1) \le p_1 \Rightarrow f(2)=2$. For $n \ge 2$, inductively and from injectivity of $f$ we have $f(p_n) > p_{n-1}$ and from Fact 6 we have $f(p_n) \le p_n$, thus is must be $f(p_n) = p_n$, for all positive integer $n$.

Now for arbitrary positive integer $n$ from Fact 4 we have
$$f(n) = f(p_1)^{\alpha_1} f(p_2)^{\alpha_2} \dots f(p_k)^{\alpha_k} = p_1^{\alpha_1} p_2^{\alpha_2} \dots p_k^{\alpha_k} = n$$
which completes our proof.