Given any infinite sequence $c_1,c_2...$ of natural numbers, if all of the natural numbers $x$ such that there exists $n$ such that $x\equiv c_n (\mod p_n)$ and $x \geq 2p$, where $p_n$ is the nth prime, are deleted from a list, must there be infinitely many natural numbers left on the list afterwards?
e.g. $c_i=1 \ \forall i\\ 1,2,3,4,\not5,6,\not7,8,\not9,\not10,\not11,12,\not13,14,\not15,\not16,\not17,18,\not19,20...$
So far I only know about the trivial case where $c_i=0$, and I may have missed obvious counterexamples.
Yes, there are infinitely many natural numbers left afterwards.
When performing Eratosthenes' unmodified sieving operation, the striking out of multiples of $2$ eliminates $\frac12$ of the values greater than $2$; the striking out of multiples of $3$ eliminates $\frac13 (1 - \frac12) = \frac16$ of the values greater than $3$; the striking out of multiples of $5$ eliminates $\frac15 (1 - \frac12 - \frac16) = \frac1{15}$ of the values greater than $5$; etc.
With your offset sieve, the numbers struck out are different, but the proportions are exactly the same. The reason is simple: each prime number is (by definition) coprime with all smaller prime numbers. Therefore there is no correlation between the numbers which are struck out before reaching that prime and the numbers which are struck out by that prime, and regardless of the value of $c_i$ we strike out $\frac1{p_i}$ of the unstruck numbers $\ge 2p_i$.
Slightly more rigorously: since the $p_i$ are distinct primes, the LCM of a set of them is simply their product, and in particular $p_i$ is coprime with $\prod_{j<i} p_j$. Then consider the equivalent classes of $\mathbb{Z} / \prod_{j\le i} p_j \mathbb{Z}$: the Chinese remainder theorem gives us a natural bijection with the Cartesian product $(\mathbb{Z} / \prod_{j\lt i} p_j \mathbb{Z}) \times (\mathbb{Z} / p_i \mathbb{Z})$. Then all elements who share a projection onto $\mathbb{Z} / \prod_{j\lt i} p_j \mathbb{Z}$ share a current status (struck out or not); and when we strike out the elements in which the projection onto $\mathbb{Z} / p_i \mathbb{Z}$ is $c_i$ we will strike out precisely $\frac1{p_i}$ of those which were not struck out (and re-strike the same proportion of those which were already struck out).
Therefore for any sequence $c_i$, and for any positive integer $C$, after striking out the values corresponding to the $j$th prime the number of unstruck numbers smaller than $C \prod_{k\le j}p_k$ will be at least as many as for the sequence $c_i = 0$ (and will not exceed that number by more than $j$, where the excess is due to numbers in $(p, 2p]$ not being struck out).