The paper Strings of congruent primes by D.K.L. Shiu contains a useful result for this problem:
Theorem 1. For each $q, a$ with $\gcd(q,a) = 1$ and large $x$, there exists a string of primes $$p_{n+1} \equiv p_{n+2} \equiv \dots \equiv p_{n+k} \equiv a \pmod q,$$ where $p_{n+k} < x$ and $$k \gg \left(\frac{\log\log x \log\log\log\log x}{(\log\log\log x)^2}\right)^{1/\phi(q)}.$$
(Here and in the rest of the answer, $p_n$ denotes the $n^{\text{th}}$ prime number.)
This is a generalization of Dirichlet's theorem on arithmetic progressions: it says that we can find arbitrarily long sequences of consecutive primes that are all contained in the progression $a,a+q,a+2q,a+3q,\dots$.
(For example, this theorem says that there are arbitrarily long sequences of consecutive primes that end in the digit $1$. A search in Mathematica says that the first time there are three consecutive primes like that is $p_{650}=4831$, $p_{651}=4861$, $p_{652}=4871$. The first time there are four consecutive primes like that is $p_{2516}=22501$, $p_{2517}=22511$, $p_{2518}=22531$, $p_{2519} = 22541$. And we can find $k$ consecutive primes like this for any $k$.)
In the prime-skipping game, let $q$ be the number of players; we will use the simpler statement that we can find $q$ consecutive primes $p_{n+1}, p_{n+2}, \dots, p_{n+q}$ that are all congruent to $1$ modulo $q$. Here is how we use this fact:
- We wait for the game to get to prime $p_{n+1}$. Call the player who gets to that prime "Alice". Alice will skip $p_{n+1}$ and say $p_{n+1}+1$, which is congruent to $2$ modulo $q$. From then on, until the next prime is skipped, all players will be locked in a cycle modulo $q$. Alice will only say numbers that are $2$ modulo $q$, the player after Alice will only say numbers that are $3$ modulo $q$, the player before Alice will only say numbers that are $1$ modulo $q$, etc.
- The next prime, $p_{n+2}$, is also $1$ modulo $q$. So the player before Alice will get to it - skipping it, and saying $p_{n+2}+1$ instead. This shifts the cycle: now the player before Alice says all the numbers that are $2$ modulo $q$, Alice says numbers that are $3$ modulo $q$, and so on; crucially, the player $2$ steps before Alice gets all the numbers $1$ modulo $q$.
- The next prime, $p_{n+3}$, is also $1$ modulo $q$. So the player $2$ steps before Alice will skip it. The cycle shifts again...
- ...and you see where this is going: the player who skips $p_{n+j}$ will be the player $j-1$ steps before Alice, for $j=1, \dots, q$. In particular, every player will get to skip one of the $q$ primes $p_{n+1}, \dots, p_{n+q}$.
Shiu's theorem also implies that there are infinitely many such strings of consecutive primes (because any particular string has a maximum length it works for, but there are arbitrarily long strings). It follows that the stronger conjecture about the prime-skipping game also holds: if we start at an arbitrary integer, eventually we will still get to a string of consecutive primes that are $1$ modulo $q$ - and that string will cause all $q$ players to skip a prime.
Note that for the primes in the string that helped us solve the prime-skipping game, the set $$\{p_{n+1} - (n+1), p_{n+2} - (n+2), \dots, p_{n+q} - (n+q)\}$$ intersects every residue class modulo $q$. Since there are infinitely many such strings, it follows that the sequence $p_n - n$ visits each residue class modulo $q$ infinitely many times. This proves the conjecture in a comment from Zhi-Wei Sun on OEIS sequence A014689.
