Let $T$ be a local minimum spanning tree that is not a global minimum.
Let $e_1,\ldots,e_m$ the edges of $T$ ordered by non-decreasing weight.
Let $T'$ be a minimum weight spanning tree that coincides with $T$ on the largest possible begin segment,
so we may assume $T'$ has edges $e_1,\ldots,e_{n-1},f_n,\ldots,f_m$ (again ordered by non-decreasing weight)
and $f_n\ne e_n$. Let $w$ be the weight function in our graph.
We claim that $w(f_n)<w(e_n)$: indeed, if $w(e_n)\leq w(f_n)$ then $e_1,\ldots,e_{n-1},e_n$ would
be a valid startup for Kruskal's algorithm and would lead to a minimum weight spanning tree that
coincides with $T$ on a larger initial segment.
$T+f_n$ contains exactly one cycle $C$. The edges of $C$ different from $f_n$ cannot all be in
$\{e_1,\ldots,e_{n-1}\}$ or we would have a cycle in $T'$. So we find at least one edge $f$ on $C$ that
is different from $e_1,\ldots,e_{n-1}$ and $f_n$. But then $w(f)\geq w(e_n)>w(f_n)$,
so $T+f_n-f$ is a spanning tree with a smaller total weight that is a neighbour of $T$ in the spanning tree graph. Contradiction.
(I left some small details for you to prove. Let me know if they cause trouble).
Note that this proves the slightly stronger statement, that each non-minimal tree has
a neighbour of strictly smaller total weight.
The cut property of the Minimum Spanning Tree (MST) problem is what you should look at. i.e. Any edge $\{x,y\}$ in an MST has weight at least as small as the edge with smallest weight in the cut that separates $x$ and $y$. This can easily be shown by contradiction.
In any cut which includes the edge $e$ (i.e. separates it's two adjacent vertices), we know that $e$ must be an edge with minimum weight in this cut.
Any spanning tree $G'$ must contain at least one edge in this cut.
Let $e'$ be such an edge. We know that
$$w(e') \ge w(e)$$
We also know that
$$w(j) \ge w(e')$$
So,
$$w(j) \ge w(e') \ge w(e)$$
Best Answer
I suppose that “the other spanning tree with the least common edge with Minimum Spanning Tree” means that if $T’$ is a spanning tree with minimum number of common edges with a given minimum spanning tree $T$. Then the claim may fail and $T’$ can be non-maximal. For instance, let $G=K_5$ be a complete graph with $n=5$ vertices. Let weight of a fixed edge $e$ of $G$ is $2$ and weights of all other edges of $G$ are $1$. Let $T$ be a spanning tree of $G$. Then $T$ is a minimum spanning tree iff $e$ is not an edge of $T$ and $T$ is a maximum spanning tree iff $e$ is an edge of $T$. To refute the claim it remains to note that there are two edge-disjoint spanning trees of $G$ not containing $e$.