I can prove that all sufficiently large integers are representable.

Firstly, observe that there is a unique way, up to isomorphism, to choose three edges $p, q, r$ of the Petersen graph such that the subgraph induced by their six endpoints is just the union of $p, q, r$.

Now, observe that we can write $2^{16}$ as the sum of semiprimes in three ways such that no prime is repeated:

$$ 101 \times 103 + 13 \times 4241 = 65536 $$
$$ 107 \times 109 + 17 \times 3169 = 65536 $$
$$ 3 \times 7 + 5 \times 13103 = 65536 $$

Label the other 12 edges of the Petersen graph with these 12 distinct primes, such that, for each edge $s \in \{p, q, r \}$, the sum (over both endpoints) of the products (over both incident edges excluding $s$) is equal to $65536$.

Moreover, do this labelling in such a way that the edges $5, 13, 17$ are incident at a vertex $v$.

Now, for each of the $2^{17}$ residue classes $R_i := \{ 2^{17}x + i : x \in \mathbb{N} \}$, we can find some $k_i \in R_i$ which is coprime to each of the aforementioned 12 primes (by Chinese Remainder Theorem). Let $K = \max \{ k_i : 0 \leq i < 2^{17} \}$ be a huge upper bound.

We're going to label each vertex with the product of the incident edges, with the exception of $v$ which will be labelled by $(5 \times 13 \times 17)k_i$ for some $i$. This means that we can attain any integer of the form:

$$ C + (5 \times 13 \times 17) k_i + 2^{16} (p + q + r) $$

for any $0 \leq i < 2^{17}$, where $C$ is a universal constant (the sum of the labels at the 3 vertices which are neither $v$ nor endpoints of $p,q,r$) and $p,q,r$ are distinct primes larger than $K$.

But by Vinogradov's theorem, *any* sufficiently large odd integer $N$ can be written as $p + q + r$ for three distinct primes larger than $K$ (because the number of representations asymptotically dominates $N^2$, so we can throw away the $O(N^2)$ representations involving repeated primes or primes beneath $K$).

So to represent an integer $I$, choose $k_i$ to make the following true:

$$ C + (5 \times 13 \times 17) k_i + 2^{16} \equiv I \mod 2^{17} $$

Then $I - C - (5 \times 13 \times 17) k_i = 2^{16} N$, where $N$ is a large odd integer expressible as the sum of three distinct primes larger than $K$. The result follows.