Assuming that $(0,1)(1,0)(0,1)^{-1}=(n,0)$, then the answer is YES. The isomophism $\phi\colon \mathbb Z[\frac1n]\rtimes\mathbb Z\to BS(1,n)$, defined by $\phi(0,1)=a$ and $\phi(1,0)=b$ and more generally
$$\phi(r,m)=a^m(a^{-q}b^pa^q)$$
where $p\in\mathbb Z$ and $q\in\mathbb Z_{\ge 0}$ such that $r=\frac{p}{n^q}$, does preserve the order.