For $>2$-ended groups. By geodesics I mean paths in the Cayley graph whose length is equal to the word metric between their initial and terminal vertex.

> If $G$ is a group with finite generating set $S$, $A$ finite and $G \setminus A$ has all components infinite and has at least three components (in the Cayley graph w.r.t. $S$). Suppose $h \notin A$, then there exists $k$ such that $A \cap kA = \emptyset$ and $kh$ is connected to $h$ in $G \setminus (A \cup kA)$.

Proof.
Consider the set $P \subset G$ consisting of all geodesics in $G \setminus A$ that begin from $\partial A$ (the boundary of $A$, i.e. elements adjacent to $A$ but not inside it), in the component of $h$. Write $P_n$ for the set of nodes first reached by some geodesic in $P$ after exactly $n$ steps. Since every translate of $A$ splits $G$ into at least three components, it is easy to show that for some $\alpha > 1$, we have $|P_n| \geq \alpha^n$ for all large enough $n$. For all $g \in P_n$, fix a geodesic representation $w_g \in S^n$, representing a the path from some element $e_g \in \partial A$ to $G$ arbitrarily.

Now pick $h'$ in $G \setminus A$ in $P_m$ for large $m$. Suppose that for all $g \in P_n$, the set $g(h')^{-1}A$ intersects the range of the geodesic $w_g$ (in the obvious sense of partial products starting from $e_n$). Then, intuitively, we can use this fact to compress the information in the paths in $P_n$. More precisely, if $g \in P_n$ then $g = g'a^{-1}h'$ for some $a \in A$ where $g'$ lies on the range of $w_g$. Because $h' \in P_m$ has word norm at least $m - |e_g|$ and $w_g$ is a geodesic, $g'$ must be in $P_{n-m+k}$ where $k \in [-3r,3r]$ where $r$ is the maximal word norm among elements of $A$. This means we can describe any element of $P_n$ by listing the at most $\lceil n/(m-3r) \rceil$ elements $a^{-1}$ and the element of $P_{n'}$ for $n' < m+3r$ obtained in the end. Now, $C |S|^{m+3r+1} |A|^{\lceil n/(m-r) \rceil}$ (where $C |S|^{m+3r}$ is an upper bound for the number of elements obtained in the end, $C$ is the boundary size of $A$) has slower asymptotic growth in $n$ than $\alpha^n$, as soon as $|A|^{1/(m-3r)} < \alpha$, a contradiction if $m$ was picked large enough.

This means we can necessarily find some $g \in P_n$ (for any sufficiently large $n$) such that $g(h')^{-1} A$ does not intersect the geodesic $w_g$. Assuming $n$ and $m$ are large enough, we have both $A \cap g(h')^{-1}A = \emptyset$, and that the connected component of $g$ in $G \setminus (A \cup g(h')^{-1}A)$ contains both $h$ and $g(h')^{-1}h$. The first fact is obvious, and for the second, observe that as $h$ and $g(h')^{-1}h$ are, respectively, in the same connected components as $h'$ and $g(h')^{-1}h' = g$ in $G \setminus A$ and $G \setminus g(h')^{-1}A$ respectively, it is suffices that $n$ and $m$ be larger than the length of a minimal path between $h$ and $h'$ in $G \setminus A$. This means we can pick $k = g(h')^{-1}$. End square.

> Let $G$ have at least $3$ ends. Then $G$ has an $n$-axial element for some $n$.

Proof.
Let $A$ finite and $G \setminus A$ has all components infinite and has at least three components, you can always find such by flood filling the finitely many finite ones. Let $h_1 \notin A$ and $h_2 \notin A$ be in distinct components of $G \setminus A$. Now, use the previous lemma to find $k_1$ such that $h_1$ and $k_1h_1$ are in the same component of $G \setminus (A \cup k_1A)$ and $A$ and $k_1A$ are disjoint. Then find $k_2$ so that $A$ and $k_2A$ are disjoint and $h_2$ is in the same component of $G \setminus (A \cup k_2A)$ as $k_2h_2$. Then $k_2^{-1}k_1$ is $n$-axial, where $n$ is bigger than the size of $A$. This follows by arguments in Cohen's thesis: Consider the sequence $..., A_{-1} = k_2A, A_0 = A, A_1 = k_1A, A_2 = k_2^{-1}k_1A...$ where you alternate left multiplication by $k_2^{-1}$ and $k_1$. Now, $(A_{-1} : A_0 : A_1)$ by assumption, and $(A_0 : A_1 : A_2)$ because if there is a path from $A_0$ to $A_2$ that does not go through $A_1$, then you can connect $k_1h_1$ to $k_1h_2$ without going through $k_1A$ (which would be a contradiction). This is because $k_1h_1$ is connected to $h_1$ and $k_1h_2$ is connected to $k_2^{-1}k_1h_2$, even in the complement of all three of those sets. By translation symmetry, $(A_i:A_{i+1}:A_{i+2})$ for all $i$, and you can show this implies $(A_i:A_j:A_k)$ for all $i < j < k$. End square.