By assumption, the points $ a_{1},\dots,a_{l} $ are distinct from $ b_{1},\dots,b_{m}. $
We know that for an affine variety $ V $ and for $ f \in k[V], $ we have the quasi-affine variety $$ V_{f} = V \backslash V(f) = \lbrace P \in V\;|\;f(P) \neq 0 \rbrace, $$ and furthermore $ V_{f} $ is isomorphic to $ \operatorname{Spec}k[V]_{f}. $
I'm not sure about the following:
So if $$ f:\mathbb{A}^{1}_{k}\backslash \lbrace a_{1},\dots,a_{l} \rbrace \longrightarrow \mathbb{A}^{1}_{k}\backslash \lbrace b_{1},\dots,b_{m} \rbrace, $$ is an isomorphism, then we have some resulting isomorphism $$ f^{*}: k[x]_{\prod_{i=1}^{l}(x-a_{i})} \longrightarrow k[x]_{\prod_{i=1}^{m}(x-b_{i})}. $$
Am I on the right track?
Let $\mathbb{K}$ be a field, let $\mathbb{A}^1_{\mathbb{K}}$ be the affine line over $\mathbb{K}$; in this case, the Zariski topology of $\mathbb{A}^1_{\mathbb{K}}$ is the cofinite topology, so any bijective and continuous function $f:\mathbb{A}^1_{\mathbb{K}}\to\mathbb{A}^1_{\mathbb{K}}$ is an homeomorphism.
Restricted $f$ to $\mathbb{A}^1_{\mathbb{K}}\setminus\{a_1,\dots,a_m\}$, one has that its image via $f$ is the (open) set $\mathbb{A}^1_{\mathbb{K}}\setminus\{b_1,\dots,b_m\}$.
In particular, this reasoning holds for any regular isomorphism $g:\mathbb{A}^1_{\mathbb{K}}\setminus\{a_1,\dots,a_l\}\to\mathbb{A}^1_{\mathbb{K}}\setminus\{b_1,\dots,b_m\}$, because $g$ is also an homeomorphism and then $l=m$!