= Solution
Take an integer $K\ge1$ bounding synchronous distances at integer times. Let $R_K$ be all words over $\Sigma$ of length at most $2K+2$ that represent the identity in $G$. This is a finite set. We claim that $\langle\Sigma\mid R_K\rangle$ presents $G$, interpreting inverse letters in the usual way.
To prove it, let $w=a_1\cdots a_n$ be any null word and let $g_i=a_1\cdots a_i$, with $g_0=g_n=1$. Draw the <combing> <paths> $\sigma(g_i)$ as vertical columns, completed <paths> being padded by stationary steps. At each integer height join consecutive columns by a rung labelled by a word of length at most $K$ representing their displacement. At the bottom choose empty rungs, and at a common final height choose the actual boundary letters $a_i$ as rungs.
Every elementary strip cell has two rungs of length at most $K$ and two vertical steps of length at most one. Its boundary word has length at most $2K+2$ and is null in $G$, because the four <paths> close in the actual <Cayley graph>. It is therefore a <relator> in $R_K$. Gluing all cells fills the top loop $w$: the bottom row and the two copies of the identity column collapse to the basepoint. Thus every null word is a product of conjugates of these finitely many short <relators>. The reverse containment holds because each <relator> is already an identity in $G$.
Hence the <finite presentation from a synchronous combing> proves
$$
\boxed{G\text{ is finitely presented}.}
$$
This is an existence proof of a finite <relator> set. It does not assume an <algorithm> for identifying identities in an arbitrary <group presentation> before establishing decidability.
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