Does the string of prime numbers $$2357111317\ldots$$ contain every natural number as its sub-string?
It follows from Dirichlet's Theorem.
If $d$ is the number we want to find, define $s=10d+1$. By definition, $\gcd(s,10)=1$ and $s$ contains the digits of $d$.
Dirichlet's Theorem's implies there's a prime of the form $p:=s+k \times 10^n$ where $10^n$ is chosen so that it has as many zeroes as digits of $s$. The digits of $d$ appear in the digits of $p$, and thus in the given string of primes.
If you mean "every non-negative integer", then the answer is yes.
First, it contains the integer 0 as a substring because 101 is a prime.
Next, for every integer i > 0, there is a prime that starts with the integer i: Take the known results about gaps between prime numbers, for example that there is always a prime between $n^3$ and$(n+1)^3$ for large n. There is always a large n such that both $n^3$ and$(n+1)^3$ start with the digits of i, therefore the prime between $n^3$ and$(n+1)^3$ also starts with the digits of i.
(This is true because one of i, 10i and 100i is not the cube of an integer).