The limiting distribution is given by the Levy's stochatic area. en.wikipedia.org/wiki/L%C3%A9vy%27s_stochastic_area A quick Google search gave me this paper that proves it. sciencedirect.com/science/article/pii/S0097316598928795

Two trivial comments: 1) In order for the question to make sense, one should consider another notion of Markov property, where both "witness" groups have decidable word problems. 2) Your reformulation with oracle is not (at least, not obviously) equivalent with an "Adian-Rabin theorem for groups with decidable word problem". Actually, neither of the implications is clear to me (I don't see an algorithm which, given a presentation of a group with decidable word problem, output an algorithm for the Word Problem)

Have you tried the search function on Kimberling's encyclopedia of triangle centers? faculty.evansville.edu/ck6/encyclopedia/Search_6_9_13.html

The lamplighter group also embed inside the Baumslag-Remneslenikov group (see arxiv.org/abs/1405.1660, it is finitely presented and metabelian). For a well-chosen generating set, the Cayley graph is the horospherical product of three “binary trees rooted at infinity” (i.e., 3-regular trees). Is this embedding quasi-isometric?

@AGenevois Do you mean the other implication is true? If the group is not finitely presented, fill in every loop of length at most L, you get a complex which is not simply connected. Take the shortest cycle which is not null-homotopic. This cycle is isometrically embedded, and has length at least L.

@SamNead Yes, the lamplighter moving along the street changing directions. Passing twice at the same place is not efficient, unless you need to switch some lamps in front of you, but your end destination is behind you. Passing three time at the same place never make sense (unless it is the starting and end point, depending how you understand « passing by »)

Here is a small observation to avoid cases: In the Diestel-Leader generating set (so $S=\{(0,1),(\delta_0,1)\}^{\pm 1}$), any geodesic can do at most 2 turnovers in the $\mathbb Z$ component, and a bi-infinite geodesic can do at most 1. Actually, up to isometries of the Cayley graph, these are the only two cases. And the geodesic with one turnover has long segments that look like geodesic of the other type. So we may suppose we work with a specific geodesic $((0,n))_{n\in\mathbb Z}$

See Lemma 3.2 in arxiv.org/abs/2305.02303 (or other references on horofunction boundaries and Busemann points).

Yes, the sequence is decreasing by the triangle inequality

See for instance arxiv.org/abs/1212.5280 for other results using $C’(\lambda)$ for $\lambda=\frac18,\frac1{10}$ and $\frac1{12}$. Their “criss-cross decomposition” might be an instance of what you’re looking for.