Sure. If you want to "disentangle" a small piece of arc (a "strand") from all the other strands around it, just declare it to be open. In other words, you can refine your solenoid by declaring every homeomorphic copy of $(0,1)$ to be an open set.

The resulting topological space is just a $\mathfrak c$-sized disjoint union of copies of $\mathbb R$. Each copy corresponds to a composant of the original solenoid.

*But wait!* you say. *You've disentangled my solenoid ***too much** -- I wanted a topology coarser than this stupid one you've described.

Well, fair enough. But I can argue that no coarser topology than the one I've described will succeed in "unwrapping" your solenoid. In other words, mine is the coarsest possible topology that refines the solenoid's original topology and is locally connected.

The reason is fairly simple: a connected and locally path connected space is path connected (here's a proof). Your disentangled solenoid should be locally arcwise connected (this is part of what I understand your word "unwrap" to mean -- furthermore, it's true of any space coarser than the one I've described). But I cannot, by refining the topology on a space, create an arc between two points that did not already have an arc between them (because an arc is compact Hausdorff, so it can't be created in a Hausdorff space by refininement). Thus the connected components of our refinement must be the path components of the solenoid. And these are precisely the composants.