Is there a sequence producing almost all primes? Is there a sequence which contains only primes and $$\lim_{n \to \infty} \frac{a_n}{p_n} = 1$$ (where $p_n$ is the $n$th prime)?

Edit: sigh. I meant a sequence produced using 0, 1, +, -, *, / used a constant amount of times, powers to a constant and any constant number of the previous elements.
Apart from the trivial solution mentioned in the comments, if you exclude a finite number $k$ of primes, then $a_n=p_{n+k}$ for $n\geq N$ for some $N$ and even in this case you have
$$ \lim_{n\to \infty} p_{n+k}/p_n=1$$
(See Cipolla's asymptotic expansion).

If you exclude primes in arithmetic progression, i.e. $a_n=p_{rn}$ for some $n$, then the limit is $r$.

On the other hand I would be surprised if the sequence $a_n$ of all primes but skipping Mersenne's primes doesn't verify what you want. Or all primes skipping twin primes. Or all primes skipping any set of rare primes.

Or if $a_n$ allow repetitions, you can find a lot more examples. For instance any set of primes in where you replace a subset of zero density of primes by the prime "2".

With the additional constraints on $a_n$ you gave, in any case solution $a_n$ to your problem has to match Cipolla's asymptotics, which are precise. I don't see how to do that with the given operations (they generate rational and exponential functions).
Primes become less common as integers become larger. This includes primes of certain forms, polynomial values and sequences. The Mersenne Numbers $2^n-1$ with $n$ prime have the condition that each divisor of $2^n-1$ must be of the form $8kn+1$ or $8kn-1$, raising the probability of them being prime compared to a random integer $N$ of similar size, however, primes of the form $2^n-1$ are still extremely rare.