A very interesting weighting is obtained by just working with directed multigraphs (dimgraphs). 7 or 8 years ago, I applied the matrix-tree theorem applied to dimgraphs in conjunction with the BEST theorem to provide a structure theory for the equilibrium thermodynamics of hybridization for oligomeric (short) DNA strands.
Briefly, the SantaLucia model of DNA hybridization takes a finite word $w$ on four letters (A, C, G, T) and associates to it various thermodynamical characteristics (e.g., free energy $\Delta G$ of hybridization) based on the sequence. Assuming the words are cyclic (which is not fair, but also not a very bad approximation practically), one has $\Delta G (w) = \sum_k \Delta G (w_kw_{k+1})$ where the index $k$ is cyclic and the 16 parameters $\Delta G(AA), \dots, \Delta G(TT)$ are given.
Assuming for convenience that the $\Delta G(\cdot,\cdot)$ are independent over $\mathbb{Q}$, it is not hard to see that $\Delta G$ projects from the space of all words of length $N$ to the space of dimgraphs on 4 vertices (again, A, C, G, T) with $N$ edges, where (e.g.) an edge from A to G corresponds to the subsequence AG. Using matrix-tree and BEST provides a functional expression for the number of words of length $N$ with a given number of AA's, AC's, ... and TT's, and thus for the desired $\Delta G$.
Going a step further, one can use generalized Euler-Maclaurin identities for evaluating sums of functions over lattice polytopes to characterize the space of all (cyclic) words of length $N$ with $\Delta G$ lying in a narrow range. This effectively completes the structure theory and shows how one can construct DNA sequences having desired thermodynamical and combinatorial properties. Among other things, I used this to design a protocol for (as David Harlan Wood put it) "simulating simulated annealing by annealing DNA".
A graph G has all spanning trees independency if and only if G does not contain two adjacent vertices v and w, neither of degree one, such that the graph G' formed by removing v and w and all their incident edges is connected. (I think this is the same as what McKay said in a comment.)
For, if G' is connected, then a spanning tree of G' plus two edges connecting it to v and w (which exist by the degree constraint) is a non-independency tree of G. And conversely, if T is a non-independency tree, let vw be an edge connecting two leaves of T; then v and w have the stated property.
As a check that this is a useful characterization (and not just a restatement of the definition): Based on this, it is trivial to test in polynomial time whether a given graph has all spanning trees independency, whereas a direct translation of the original definition into an algorithm would be exponential.
ETA: In response to McKay's request for a more structural characterization, here's a description of the graphs with all trees independency, rather than (as above) of the complementary class of graphs. Given a graph G, partition G into its biconnected components, and partition each nontrivial biconnected component into its SPQR tree (system of triconnected components). Then G has all trees independency if and only if, for every non-virtual edge of an R or S node of one of these SPQR trees, at least one endpoint is an articulation vertex. (This is because the edges that do not separate two triconnected components are exactly these edges, so to avoid leaving G' connected they have to instead separate two biconnected components.) And again, this is an improved characterization and not just a restatement of the earlier characterization, because it gives a linear time algorithm while a naive implementation of the earlier characterization would be a higher polynomial.
Best Answer
I think that conjecture is true. Let $a$, $b$ be two non-adjacent vertices. I claim that $H = G \setminus \{a, b\}$ must contain exactly two connected components. Indeed, if it has more than two, we can take any of these, say $H_1$, and $H_1 \cup \{a, b\}$ will be connected while the rest of $G$ not. On the other hand, the case of a single component $H$ is also impossible since otherwise $G \setminus H$ is just pair of vertices $a$ and $b$ and it is disconnected. Thus, $G$ consists of two components $H_1$ and $H_2$ and $a$ and $b$ in between.
Now I claim that $H_1$ is just a path from $a$ to $b$. First, since $G \setminus H_2$ is connected, there should be some path $l$ from $a$ to $b$ which lies entirely in $H_1$. On the other hand, $H_1$ cannot contain any other vertices since otherwise $G \setminus \{l \cup a \cup b \}$ is disconnected. The same holds for $H_2$. Thus, $G$ originally was just a cycle.