You have given $L$ as a general linear operator between Hilbert spaces $H_1$ to $H_2$ (the one indexed by $j$ and $i$ respectively). That is all possible linear operators, the entirety of $Mor(H_1,H_2)$. Mor here is short for the word morphism, a word for whatever the notion of whatever sorts of functions/operators/etc that are allowed. In the present context of vector spaces, this just means what are the linear operators $H_1 \to H_2$.
So we have all of the linear operators and for each of them $\mathbf{vec}$ produces a pure state in $H_2 \otimes H_1$. There is nothing missing. There are no other linear operators. There is no way to get states $\rho$ on $H_2 \otimes H_1$ that aren't pure here. Only mixed.
You already noticed this is invertible. That is we have $Mor(H_1,H_2) \simeq H_2 \otimes H_1$. The LHS is where $L$ is an element and the RHS is where $\mathbf{vec}(L)$ is an element.
There are extensions of this concept. For example suppose we have 3 Hilbert spaces. $H_1$, $H_2$ and $H_3$. Giving a linear map from $H_1$ to the space of linear maps from $H_2$ to $H_3$, which would be denoted as $Mor(H_1, Mor(H_2,H_3))$ is the same as giving a linear operator from $H_1 \otimes H_2 \to H_3$. See how this implies the above. Take $H_3=\mathbb{C}$. Then a way of thinking about a bra for $H_2$ is as a linear map $H_2 \to H_3=\mathbb{C}$. I'll leave you to ponder this fact some more to fully flush out how it implies the above.