Here I have a conjecture:
Let $G$ be a simple undirected graph. The neccessary and sufficient condition for a graph $H, L(H) = G$ to exist is to have no vertex $u \in V(G)$ so at least three adjacent vertices of $u$ appear in an independent set.
It surely is a necessary condition:
Let's assume that there is a vertex $u \in V(G)$ with three adjacent vertices in an independent set, and let $xy$ be its corresponding edge in $H$. According to the pigeonhole principle, there are at least two edges $e_1, e_2$ that their corresponding vertices in $G$ are independent, and are adjacent with $xy$ in $H$, where their common vertex is either $x$ or $y$. In this case, $e_1, e_2$ are adjacent, that contradicts with the independence of their corresponding vertices in $G$.
Your condition is equivalent to forbidding $K_{1, 3}$ as an induced subgraph (the 3 branches star graph). However, this is not sufficient to characterize fully the line graphs. However, a characterization exists by excluding induced subgraphs, which requires eight more forbidden subgraphs. These eight graphs provide the minimal counterexamples to your conjecture.
Another (easy) characterization is that a line graph needs to admit a clique partition where each vertex belongs to exactly two cliques.