Well, in the absence of any answers, perhaps this might help somebody to get a proper solution.
In order to show that there are infinitely many composite pairs of the form $n!\pm1$, it would suffice to prove that the expected number of prime numbers of the form $n!\pm1$ is
relatively small, i.e.
$$\limsup\limits_{N\to\infty}\frac{E|\{n=1,\dots,N|\ n!+1\ \mbox{or } n!-1\ \mbox{is prime}\}|}{N}=0.$$
Now, there is a note by Caldwell and Gallot (who were mentioned in Kevin Buzzard's comment avove) which contains a non-rigorous probabilistic argument yielding a heuristic estimate of the expectation.
In short, they start with a rough assumption that $n!\pm1$ behaves like a random variable
and use the Stirling formula $\log n!\sim n(\log n-1)$. The prime number theorem shows that the probability of a random number of the size $\sim n!\pm1$ being prime is
$$P_n\sim\frac{1}{n(\log n-1)},\quad n\gg 1. $$
Then they take into account Wilson's theorem and some other obvious obstacles to $n!\pm1$ behaving randomly, and obtain just a slightly weaker estimate
$$P_n\sim\left(1-\frac{1}{4\log 2n}\right)\frac{e^\gamma}{n}$$
where $γ$ is the Euler–Mascheroni constant. The latter estimate translates into the estimate of the expected number of factorial primes of each of the forms $n!\pm1$, $n\leq N$
$$E_N\sim e^\gamma \log N,\quad N\gg 1.$$
Now, this is actually more than we need, and hopefully the probabilistic argument can be made rigorous to show that $E_N/N$ goes to $0$ as $N\to\infty$.
Edit added.
Is it true that for every positive integer $B$ there is a positive integer $N$ such that $N$ is divisible by all primes up to $B$, and $N \pm 1$ are both composite?
The modified question is easy. Take $N=(B!)^3$.
As long as you axiomatize set theory in first-order logic, the answer to your question is no. The axioms would be consistent with each finite subset of the following set of sentences involving a new constant symbol $c$: "$c$ is a natural number" and "$c\neq n$" for each (standard name of a) natural number $n$. By compactness, there would be a model of the axioms plus all of these sentences, and in that model $c$ would denote a nonstandard natural number.
On the other hand, if you're willing to go beyond first-order logic, then the answer to your question is yes. For example, in second-order logic, you can express the induction axiom as a single sentence and be confident that "set" really means arbitrary set (not "internal set" or anything like that). In other words, once you're sure that "set" has its intended meaning, the induction principle guarantees that "natural number" also has its intended meaning. (To me, this doesn't look very helpful, since the intended meaning of "set" seems more complicated than the intended meaning of "natural number".)
For another example, if you're willing to use infinitary logic, then you can formulate the axiom "every natural number is equal to 0 or to 1 or to 2 or to 3, or ..."
Best Answer
The answer is no.
Proof: Let $\phi(u,v)$ be a $\Sigma_1$ formula representing $f$. The formula $$\psi(w)=\exists x\in[a,b]\,\forall u<w\,\forall v\,(\mathrm{Prime}(u)\land\phi(u,v)\to x\equiv v\pmod u)$$ holds for all standard $w$, hence it holds for some nonstandard $w$ by overspill. QED
Proof: Use $f(p)=0$ for 1, and e.g. $f(p_n)=n$ for 2. QED
In fact, we do not need anything as strong as PA for the argument to work. Recall that $E_1$ is the class of all bounded existential formulas.
Rather than giving a direct proof, let me mention that this is an immediate consequence of the following more general result:
On the other hand, the situation is different for theories on the weaker side of the Tennenbaum barrier.
In fact, it is easy to see that this holds in the Shepherdson model, consisting of Puiseux polynomials $$a_nx^{n/m}+a_{n-1}x^{(n-1)/m}+\dots+a_0,$$ where $a_i\in\mathbb R$, $a_0\in\mathbb Z$: namely, every such polynomial with $a_0=0$ is divisible by all standard integers.
Reference:
[1] George Wilmers, Bounded existential induction, Journal of Symbolic Logic 50 (1985), no. 1, pp. 72–90. JSTOR