Claim: Yes, that statement is true.

Proof Sketch: Let $$T_1,T_2$$ be two minimal spanning trees with edge-weight multisets $$W_1,W_2$$. Assume $$W_1 \neq W_2$$ and denote their symmetric difference with $$W = W_1 \mathop{\Delta} W_2$$.

Choose edge $$e \in T_1 \mathop{\Delta} T_2$$ with $$w(e) = \min W$$, that is $$e$$ is an edge that occurs in only one of the trees and has minimum disagreeing weight. Such an edge, that is in particular $$e \in T_1 \mathop{\Delta} T_2$$, always exists: clearly, not all edges of weight $$\min W$$ can be in both trees, otherwise $$\min W \notin W$$. W.l.o.g. let $$e \in T_1$$ and assume $$T_1$$ has more edges of weight $$\min W$$ than $$T_2$$.

Now consider all edges in $$T_2$$ that are also in the cut $$C_{T_1}(e)$$ that is induced by $$e$$ in $$T_1$$. If there is an edge $$e'$$ in there that has the same weight as $$e$$, update $$T_1$$ by using $$e'$$ instead of $$e$$; note that the new tree is still a minimal spanning tree with the same edge-weight multiset as $$T_1$$. We iterate this argument, shrinking $$W$$ by two elements and thereby removing one edge from the set of candidates for $$e$$ in every step. Therefore, we get after finitely many steps to a setting where all edges in $$T_2 \cap C_{T_1}(e)$$ (where $$T_1$$ is the updated version) have weights other than $$g(e)$$.

Now we can always choose $$e' \in C_{T_1}(e) \cap T_2$$ such that we can swap $$e$$ and $$e'$$¹, that is we can create a new spanning tree

$$\qquad \displaystyle T_3 = \begin{cases} (T_1 \setminus \{e\}) \cup \{e'\} &, w(e') \lt w(e) \\[.5em] (T_2 \setminus \{e'\}) \cup \{e\} &, w(e') \gt w(e) \end{cases}$$

which has smaller weight than $$T_1$$ and $$T_2$$; this contradicts the choice of $$T_1,T_2$$ as minimal spanning trees. Therefore, $$W_1 = W_2$$.

1. The nodes incident of $$e$$ are in $$T_2$$ connected by a path $$P$$; $$e'$$ is the unique edge in $$P \cap C_{T_1}(e)$$.