 Solution

Assume first that deleting an edge leaves two transient graph components. One component supports a finite-energy unit flow from its endpoint to infinity by the finite-energy flow criterion for transience. Extend that flow by zero across the deleted edge and throughout the other component. It remains a finite-energy unit flow on the whole tree, so the tree is a transient graph.

Conversely, root a transient tree at $o$ and take its finite-energy unit electrical flow to infinity. On a tree this flow can be chosen nonnegative in every direction away from the root: it is obtained as the limit of the currents to wired boundaries, each of which sends nonnegative current into descendant subtrees. At each nonroot vertex its incoming current equals the sum of its outgoing currents, by flow conservation.

If every vertex reached by positive current had exactly one positive-current child, the unit current would run along a single infinite ray. Every edge of that ray would carry current one, so its energy of a flow would be ${\sum }_{j\ge 1}1=\infty$, a contradiction. Thus there is a vertex $v$ with two positive-current child edges, say $(v,x)$ and $(v,y)$.

The restricted flow below $x$, rescaled by its incoming current, is a finite-energy unit flow from $x$ to infinity; hence the component below $x$ is a transient graph. After removing $(v,x)$, the other component contains $v$, the edge $(v,y)$, and the descendant subtree below $y$. Send unit current along $(v,y)$ and then use the rescaled descendant flow below $y$, with zero current elsewhere. Its energy of a flow is finite, so this component is a transient graph too. ThereforeUnit edge resistances are essential here: an infinite ray with sufficiently rapidly decreasing resistances can be transient without any such splitting edge.