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
If $G$ contains a cycle of edges each of the same minimum weight, then it is not possible for $G$ to have a spanning tree without omitting at least one of those edges.
Otherwise the answer is yes: Let $T$ be a spanning tree of $G$ NOT containing $e$. Then $T+e$ contains one cycle. By assumption, there is an edge in the cycle of non-minimum weight. Remove that edge, and we have a new spanning tree $T'$ of lower weight.