If we take the first $k$ primes, are there always going to be natural numbers not representable as:
$$p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k} \pm p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}\ ?$$
Assume non-negative $a_i$, in case that matters.
e.g. It seems to me as though no $2^a3^b \pm 2^c3^d=11^2$.
On
This is immediate from results on $S$-unit equations. Indeed, let $p_{k+1}$ be the next prime after $p_1,\dots,p_k$. Any solution to $$p_1^{a_1}\dots p_k^{a_k}\pm p_1^{b_1}\dots p_k^{b_k}=p_{k+1}^c,$$ upon division by $p_{k+1}^c$, gives a different solution to the $S$-unit equation for $S=\{p_1,\dots,p_{k+1}\}$. General theory implies that there are only finitely many such solutions, hence in particular there are only finitely many $c$ for which a solution exists.
There are extensive tables which list all solutions of the $S$-unit equation for some small $S$, see for instance here. Looking at the fifth file and for solutions with radical $22,33,66$ (there are none for the former two, and for the latter there are a couple, but none involving $11^2$) we see that there indeed isn't a solution to $2^a3^b\pm 2^c3^d=11^2$.
The answer to your question seems to be that, for the first $k$ primes, there'll always be natural numbers which are not representable as
$$p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k} \pm p_1^{b_1} p_2^{b_2} \cdots p_k^{b_k} \tag{1}\label{eq1}$$
The journal Archiv der Mathematik, published by SpringerLink since $1948$, has an article in June $2012$, Volume $98$, Issue $6$, pp $527$–$533$, with an introductory page at Representing integers as linear combinations of power products. It gives a preview, with the introduction starting with
This indicates that even a considerably more general result than what you asked about holds, i.e., not all positive integers can be represented by any fixed number of sums and/or differences of elements of a finite set of powers of primes.