I'll write $T_f$ for the set of primes $p$ such that $f(x)$ is a bijection $\mathbb{F}_p \to \mathbb{F}_p$. I claim that $T_f$ is always either finite or else $\# \{ p \in T_f : p \leq y \} \sim c \tfrac{y}{\log y}$ for some $c>0$.
The following result is known as Schur's conjecture; a flawed proof was given by Fried, with corrected versions by Turnwald and Müller:
Theorem Let $f(x) \in \mathbb{Z}[x]$. If $T_f$ is infinite, then $f(x)$ is a composition of linear polynomials and Dickson polynomials.
One should note that the Dickson polynomial $D_n(x,0)$ is just $x^n$, so this includes the possibility of including monomials in our composition.
A composition of functions $\mathbb{F}_p \to \mathbb{F}_p$ will be bijective if and only if all the functions composed are bijective. Linear functions are always bijective; the monomial $D_n(x,0)=x^n$ is bijective iff $GCD(n,p-1)=1$ and, for $a \neq 0$, the Dickson polynomial $D_n(x,a)$ is bijective iff $GCD(n,p^2-1)=1$ or, equivalently, $GCD(n,p-1) = GCD(n,p+1)=1$. (This last statement is copied from Lemma 1.4 in Turnwald; I didn't check it.) In short, imposing that our composition is bijective imposes finitely many conditions on the residue class of $p$ modulo various integers.
If these modular conditions can be satisfied by infinitely many primes, then they are satisfied by $\sim c \tfrac{y}{\log y}$ primes $\leq y$, by the PNT in arithmetic progressions.
Polynomials where $T_f$ is infinite are exceptional polynomials. (The definition of "exceptional" is that there are infinitely many prime powers $q$ such that $f$ is bijective on $\mathbb{F}_q$, so it also allows examples like $x^2$ which is bijective on $\mathbb{F}_{2^k}$ but not on $\mathbb{F}_p$ for any odd $p$; imposing bijectivity for infinitely many primes rather than prime powers is obviously even more restrictive.)
As the name suggests, exceptional polynomials are very rare for degree $\geq 5$. For example, they all have solvable Galois group, where as almost all degree $n$ polynomials have Galois group $S_n$. Also, by completing the square, quadratics won't have $T_f$ infinite, although they might be exceptional because of $\mathbb{F}_{2^k}$. I'm not sure what happens in degrees $3$ and $4$.