If $(a_n)\to A\neq 0$ and $(a_n b_n)\to AB$ then $(b_n)\to B$

Your proof is correct but I'd recommend justifying that $a_n \neq 0$ for all $n$ sufficiently large. Note that this is necessary to make sense of the expression $$ b_n = \left( \frac{a_nb_n}{a_n} \right). $$ To justify that $a_n \neq 0$ for all $n$ sufficiently large, we will use that $A = \lim{a_n} \neq 0$. Because $a_n \to A$, there exists $N \in \mathbb{N}$ such that $$ |a_n - A| < \epsilon = \frac{|A|}{2} $$ for all $n \geq N$. Therefore, for all such $n$, \begin{align*} |A| - |a_n| \leq \left\vert a_n - A \right\vert< \frac{|A|}{2} \end{align*} whence $$ 0 < \frac{|A|}{2} < |a_n|, \quad \forall n\geq N. $$ This means that the $N$-tail of $(a_n)$ is never zero and consequently the sequence $$ \frac{b_n}{a_n} $$ is well defined for all $n \geq N$ (and we only really care about the tail of a sequence when discussing limits). Then, by invoking the theorem and lemma you've cited, it follows that \begin{align*} \lim{b_n} = \lim\left(\frac{a_n b_n}{a_n} \right) = \frac{\lim(a_nb_n)}{\lim{a_n}} = \frac{AB}{A} = B. \end{align*}