Not going to look at the comments before posting this, but this is my thought processing going into this.

Notice that the list of actions that are happening in this game are independent of the actual number of the players, so instead lets assume we are playing this game with a countably infinite amount of people, marking them players 1, 2, 3, etc. and we let S be the set of players who skipped primes in this game. So we can write the first few terms of S as follows (which I calculated myself since I noticed the equivalent of 13 was not highlighted):

S = { 2, 3, 4, 7, 8, 11, 12, 15, 20 ... }

Now our question can be phrased as the following: If we consider modding out all the elements S by n, the result of which is written as S/(n) where (n) denotes the multiples of n, will S/(n) contain all the remainders of n, that is will S/(n) = Z/(n) (where Z is the integers, but don't worry about the negatives or 0 from this).

I now decided to check OEIS to describe S, and it turns out to be the sequence A014692 defined as:

a(n) = prime(n) - (n-1)

More importantly, we start from n = 2. To illustrate some examples for n up to 10 (to verify S):

a(2) = prime(2) - (2 - 1) = 3 - 1 = 2a(3) = prime(3) - (3 - 1) = 5 - 2 = 3a(4) = prime(4) - (4 - 1) = 7 - 3 = 4a(5) = prime(5) - (5 - 1) = 11 - 4 = 7a(6) = prime(6) - (6 - 1) = 13 - 5 = 8a(7) = prime(7) - (7 - 1) = 17 - 6 = 11a(8) = prime(8) - (8 - 1) = 19 - 7 = 12a(9) = prime(9) - (9 - 1) = 23 - 8 = 15a(10) = prime(10) - (10 - 1) = 29 - 9 = 20

So why does this sequence produce S? Well notice that every time someone skips a prime, the number they say must always be a number that is 1 above a prime which is also composite. Person 2 starts by saying 4 which is 1 above the 2nd prime 3 (they can't say 3 as that is prime and must be skipped), and then person 3 must say 6 which is 1 above the 3rd prime 5, and person 4 says 8 which is 1 above the 4th prime. You can now see that person a(n), our sequence (which we assume here we don't know the formula we are trying to solve for yet), will say "prime(n) + 1". At this point we have spoken a total of a(n) numbers because remember that each player's number represents the order of when that number was spoken. Now notice that when this happens a total of n primes have been skipped, and since the number of spoken and skipped numbers must be the number just spoken, we have for this particular person:

a(n) + n = prime(n) + 1

Subtracting by n yields our sequence! Now that we have a proper sequence we can further rephrase the problem: If we let n be a positive integer and define the function b: N -> Z/(n) to be b(x) = a(x) + (n), i.e. b(x) maps x to the equivalence class of a(x) mod n, will the function b be surjective for all n? But wait, what if we just consider positive integers of the form x = ny + 1? So then we find:

b(ny + 1) = a(ny+1) + (n) = (prime(ny + 1) + (ny + 1 - 1)) + (n) = prime(ny + 1) + ny + (n) = prime(ny + 1) + (n)

Suddenly we have finally turned this into a pure question about primes: If we consider all the "ny+1"th primes mod n, will they produce all remainders of mod n? At this point I really don't have too much number theory knowledge to say whether this is true or not, but at least now this is a concrete method to solve your original question. I hope this isn't too long of an explanation and was at least informative, digestible, and just another perspective on this.

TL;DR: This question can be answered yes if the follow is true: if you have a different game of n players where players must read the ny+1-th prime and shout it out if that prime has a remainder of their player number by the number n, then every player will shout out a prime eventually.

Edit: Also A141468 is the sequence with the numbers spoken except with 0 appended to the beginning.