First, following your answer in comments, I assume that $(0,1)(1,0)(0,1)^{-1}=(n,0)$, so that the map $\phi\colon \mathbb Z[\frac1n]\rtimes\mathbb Z\to BS(1,n)$, 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 $r=\frac{p}{n^q}$ is an isomorphism.

Then the order you propose on $BS(1,n)$ is not well-defined. There are two issues:

- First $b=a^{-1}b^na$, so if I follow correctly you would declare $b<a^{-1}b^na$, despite the two elements being equal (here I assume $n\ge 2$).
- Second one has $a^{-1}ba=a^{-2}b^na^2$. If $n<0$, you would declare this element both larger (first expression) and smaller (second expression) than $e$.

In order to define the order on $BS(1,n)$, the easiest is to just define when an element is larger than $e$. In this case, one define $a^m(a^{-q}b^pa^q)>e$ if $m>0$ or ($m=0$, $q$ is even and $p>0$). We can then extends this order to get a right-invariant order: $g>h$ iff $gh^{-1}>e$. This coincides with the "push-forward" (via $\phi$) of the right-order defined on $\mathbb Z[\frac1n]\rtimes\mathbb Z$, i.e., the answer is "Yes".

(If you want to directly define the order on $BS(1,n)$, you need to force $m_1=m_2$ to avoid the first problem, and be careful about the parity of $m_1,m_2,n_1,n_2$ to avoid the second.)