Let $G=\mathbb{Z}\times\mathbb{Z}/7\mathbb{Z}$. Let $S$ consist of $(1,1),(1,0),(0,1)$ and their inverses. Let $x=(0,3)$.

It seems clear that the neighbors of $x$ are contained in $\{-1,0,1\}\times\{2,3,4\}$, since each coordinate of a neighbor of $x$ will differ by $0,1$, or $-1$ from the corresponding coordinate of $x$.

If $(n,k)\in G$ and $n>7$, then it is clear that the word norm $|(n,k)|_S$ is equal to $n$. Hence, any geodesic word representing $(n,k)$ uses only the generators $(1,1)$ and $(1,0)$. Similarly, if $n<-7$, and geodesic word representing $(n,k)$ uses only the generators $(-1,-1)$ and $(-1,0)$. It follows that $J_S$ contains no neighbors of $x$, as

$J_S\cap (\{-1,0,1\}\times\mathbb{Z}/7\mathbb{Z})=\{(-1,-1),(-1,0),(0,0),(1,0),(1,1)\}$.

Thus, $d(x,J_S)\geq 2$. But $|x|_S=3$.

Let $G=\mathbb{Z}\times\mathbb{Z}/7\mathbb{Z}$. Let $S$ consist of $(1,1),(1,0),(0,1)$ and their inverses. Let $x=(0,3)$.

It seems clear that the neighbors of $x$ are contained in $\{-1,0,1\}\times\{2,3,4\}$, since each coordinate of a neighbor of $x$ will differ by $0,1$, or $-1$ from the corresponding coordinate of $x$.

If $(n,k)\in G$ and $n>7$, then it is clear that the word norm $|(n,k)|_S$ is equal to $n$ (representing $k$ by some element of $\{0,1,\ldots, 6\}$, we have that $a^kb^{n-k}$ represents $(n,k)$, where $a=(1,1)$ and $b=(1,0)$, and it is obvious that no shorter word can represent $(n,k)$). Hence, any geodesic word representing $(n,k)$ uses only the generators $(1,1)$ and $(1,0)$. Similarly, if $n<-7$, any geodesic word representing $(n,k)$ uses only the generators $(-1,-1)$ and $(-1,0)$. It follows that $J_S$ contains no neighbors of $x$, as

$J_S\cap (\{-1,0,1\}\times\mathbb{Z}/7\mathbb{Z})=\{(-1,-1),(-1,0),(0,0),(1,0),(1,1)\}$.

Thus, $d(x,J_S)\geq 2$. But $|x|_S=3$.