Q1. $p$ is $\star$-prime iff equation $xy+x+y=2p$ has no solution and $xy+x+y=2p-1$ has exactly one solution, i.e. $(x+1)(y+1)=2p+1$ has no solution (which is iff $2p+1$ is prime) and $(x+1)(y+1)=2p$ has only one solution $\{x,y\}=\{1,p-1\}$. This last holds iff $p$ is prime.
Q2. Why partitions?! The number of $\star$-divisors of $n$ equals $\tau(2n+1)+\tau(2n)-4$.
To give a vague answer, I think these questions are difficult because they mix multiplicative conditions (being prime) and additive conditions (as in the twin prime case).
Basically all results about primes that I can think of come down to unique factorization of the integers. For example, the zeta function is given as
$$\zeta(s) = \sum_n n^{-s} = \prod_p (1 - p^{-s})^{-1}.$$
The right hand side is why the zeta function tells you about prime numbers, but the left hand side is what typically helps you prove theorems. For example, Riemann noticed that the left side looked like something similar to what Poisson summation is good for, and hence proved analytic continuation and the functional equation.
On a simpler level, one nice proof that there are infinitely many primes is to observe that $\sum_n 1/n$ diverges, by elementary calculus, and therefore the right hand side diverges for $s = 1$ as well.
Gerhard Paseman suggested looking at arithmetic progressions, and I think this is an extremely instructive example. Looking at the sum of $n^{-s}$ restricted to an arithmetic progression, you don't have any equation like the above. Conversely, if you take a product over only the primes $p$ in some arithmetic progression, you don't get anything nice like the left side. However, if you let $\chi$ be a Dirichlet character, e.g., a homomorphism from $(\mathbb{Z}/N)^{\times}$ to $\mathbb{C}$, then you get the Dirichlet $L$-function
$$L(s, \chi) = \sum_n \chi(n) n^{-s} = \prod_p (1 - \chi(p) p^{-s})^{-1}.$$
In some way this is forcing a round peg into a square hole: the arithmetic progression condition couldn't be handled directly. But it can be written as a linear combination of Dirichlet characters, and once you force everything to be multiplicative, the machinery (Poisson summation, etc.) all works.
So in other words, IMHO, the question isn't "why is the twin prime conjecture difficult", but "why can we prove anything about the primes at all?" Our toolbox is, in my experience, still pretty limited.
Best Answer
Here is a general result. For a sequence of nonnegative numbers $\{a_n\}$, let $A(x) = \sum_{n \leq x} a_n$. For example, if $S \subset \mathbf Z^+$ and we set $a_n = 1$ when $n\in S$ and $a_n = 0$ when $n \not\in S$, then $A(x)$ is the number of elements of $S$ that are $\leq x$.
Exercise: If $A(x) = O(x/(\log x)^r)$ for a positive integer $r$ and all $x \geq 2$, then $\sum_{n \leq x} a_n/n$ converges as $x \to \infty$ if $r \geq 2$ and $\sum_{n \leq x} a_n/n = O(\log \log x)$ for $r = 1$.
Example: if $f_1(T), \ldots, f_r(T)$ are polynomials with integer coefficients that fit the hypotheses of the Bateman-Horn conjecture (twin primes are $f_1(T) = T$ and $f_2(T) = T+2$, while Sophie Germain primes are $f_1(T) = T$ and $f_2(T) = 2T+1$), then Bateman and Stemmler showed $60$ years ago that the number of $n \leq x$ such that $f_1(n), \ldots, f_r(n)$ are all prime is $O(x/(\log x)^r)$, where the $O$-constant depends on $f_1, \ldots, f_r$. Therefore if above we take $S$ to be the $n \in \mathbf Z^+$ such that $f_1(n), \ldots, f_r(n)$ are all prime and define $a_n$ to be $1$ or $0$ according to $n \in S$ or $n \not\in S$, then the exercise above says the sum of all $1/n$ for $n \in S$ converges if $r \geq 2$.
So for any sequence of pairs of primes $p$ and $ap+b$ that are expected to occur infinitely often ($p$ and $p+2$, or $p$ and $2p+1$, or $\ldots$), the sum of $1/p$ for such primes converges.
That the sum of the reciprocals of the twin primes converges indicates that this summation is the wrong thing to be looking at. We want a strategy to prove the infinitude of twin primes, and that suggests a better sum. The Bateman-Horn conjecture predicts the number of $n \leq x$ such that $f_1(n), \ldots, f_r(n)$ are all prime is asymptotic to $Cx/(\log x)^r$ where $C$ is a positive constant depending on $f_1, \ldots, f_r$, and if $A(x) \sim cx/(\log x)^r$ as $x \to \infty$ for some $c > 0$ then $\sum_{n \leq x} a_n(\log n)^{r-1}/n \sim c\log\log x$. Therefore we expect (but have never proved) that the sum of $(\log p)/p$ over prime $p \leq x$ such that $p$ and $p+2$ are prime should grow like $c\log\log x$ for some constant $c > 0$, and a similar asymptotic estimate (for a different constant $c$) should hold for the sum of $(\log p)/p$ over all prime $p \leq x$ such that $p$ and $2p+1$ are prime.