The conjecture is true if you don't insist that the demi-primes are consecutive. Choose distinct large primes $q_S$ indexed by the 2-element subsets $S$ of $G$. It suffices to construct odd primes $p_x$ for $x \in G$ such that
$p_x$ is not equal to $q_S$ for any $x$ and $S$
$(p_x+1)/2$ is divisible by $q_S$ if and only if $x \in S$ and the edge $S$ is not part of $G$, and
for distinct $x$ and $y$, the numbers $(p_x+1)/2$ and $(p_y+1)/2$ have no common prime factors except for possibly the $q_S$.
In fact, this is easy: just choose the $p_x$ one at a time. The conditions imposed on any one $p_x$ by 2) and 3) amount to finitely many congruence conditions with prime moduli, and each prime modulus appears at most once, and none are asking $p_x$ to be divisible by the modulus, so they are satisfiable by Dirichlet's theorem on primes in arithmetic progressions.