About the prime divisors of values of polynomials

We will assume that $f$ is irreducible over $\mathbb{Q}$. Let $\gamma$ be a root of $f$, and let $K = \mathbb{Q}(\gamma)$. By Chebotarv's density theorem, the set $P_f$ of primes $p$ with the property that $f(x) \equiv 0 \pmod{p}$ has at least one root in $\mathbb{F}_p$ has positive relative density in the set of rational primes, and the set of integers with at least one prime factor in $P_f$ has positive density. For each prime in $P_f$, we can solve the congruence $f(x) \equiv 0 \pmod{p}$ in the integers; hence $P_f$ is precisely the set of primes you are looking for. Since $P_f$ has positive relative density in the primes, the sum of their reciprocals will diverge, with the constant as predicted by Chebotarv's density theorem.

Assume without loss of generality that $P$ is irreducible, and denote by $S_P(X)$ the set of primes $p < X$ that divide some value $P(n)$. Let $G$ be the Galois group of $P$ and $n_1 > 0$ the number of its elements that have at least one fixed point.

Then, as Noam Elkies says, $$ \sum_{p \in S_P(X)} \frac{1}{p} = \frac{n_1}{|G|} \, \log{\log{X}} + \mathrm{const} + O_P\Big( 1 \big/ \log{X} \Big). $$ Here is a sketch of a simple proof in the case that $P$ is Galois (so that $n_1 = 1$ and $|G| = \deg{P}$), yielding the lower bound $|G|^{-1} \log{\log{X}} + O_P(1)$ for the general case.

By partial summation (assume $P$ is Galois here...), this is equivalent to $$ \sum_{p \in S_P(X)} \frac{\log{p}}{p} = \frac{1}{\deg{P}} \, \log{X} + O_P(1). $$ Adapt the argument for the variant of Selberg's proof I outlined here of Dirichlet's theorem: Shortest/Most elegant proof for $L(1,\chi)\neq 0$, replacing $\mathbb{Z}[\zeta_N]$ with $A = \mathbb{Z}[\alpha]$ and $V(X)$ by $\prod_{n_i < X^{1/d}} (n_1\alpha^0 + \cdots + n_d \alpha^{d-1})$, where $d := \deg{P}$ and $\alpha \in \mathbb{C}$ is a root of $P$ multiplied (to clear denominator) by the leading coefficient of $P$.

The key point is that the non-zero ideals of $A$, viewed as lattices in the real vector space $A \otimes \mathbb{R}$ and then scaled to have unit volume, belong to a compact subset of the space of unimodular lattices. This is not too hard to see, and can be used by a standard argument to prove that every non-zero ideal $I \subset A$ contains $X/\mathcal{N}(I) + O_P\Big( (X/\mathcal{N}(I))^{1-\frac{1}{d}} \Big)$ of the factors from $V(X)$. (The exponent $1-1/d$ in the error term here is the same as the "trivial" exponent in the $d$-dimensional extension of Gauss's circle problem.) Since $\sum_{p < X} (\log{p})/p^2$ and $\frac{1}{X} \sum_{p < X} (X/p)^{1-\frac{1}{d}}\log{p}$ are bounded as $X \to \infty$, while it is easy to evaluate $\frac{1}{X} \log{|V(X)|} = \frac{1}{d} \log{X} + O_P(1)$, we may conclude a la Chebyshev and Mertens the desired estimate $\sum_{p \in S_P(X)} \frac{\log{p}}{p} = \frac{1}{d} \log{X} + O_P(1)$. (The assumption that $P$ is Galois is used in the relation of rational primes to prime ideals: each sufficiently big prime $p \in S_P(X)$ is contained by exactly $d$ degree-one prime ideals.)

Nonetheless, showing that $\zeta_K(s)$ has a simple pole at $s=1$ and following GH's answer is probably easier than verifying the two claims in my previous paragraph.

I don't think one needs to invoke the Chebotareev Density Theorem here.

Assume that the given sum converges, and let $K$ be a number field generated by any root of $P(x)$. Then, by assumption, the sum of $\mathrm{Norm}(\mathfrak{p})^{-1}$ over the degree $1$ prime ideals $\mathfrak{p}$ in $K$ converges, hence the same holds for the sum of $\mathrm{Norm}(\mathfrak{p})^{-1}$ over all prime ideals $\mathfrak{p}$ in $K$. This implies that the Dedekind zeta function of $K$, $$ \zeta_K(s)=\prod_{\mathfrak{p}}\frac{1}{1-\mathrm{Norm}(\mathfrak{p})^{-s}},\qquad \Re(s)>1, $$ tends to a finite limit as $s\to 1+$, while it is known that $\zeta_K(s)$ has a simple pole at $s=1$.