We know:
$$\mathbb{P}[A|B]=\frac{\mathbb{P}[A \wedge B]}{\mathbb{P}[B]}$$
$$\mathbb{P}[A|\neg B]=\frac{\mathbb{P}[A\wedge\neg B]}{\mathbb{P}[\neg B]}$$
We also know
$$1-\mathbb{P}[B]=\mathbb{P}[\neg B]$$
My question is, can we obtain, or is there another simpler relation between $\mathbb{P}[A|B]$ and $\mathbb{P}[A|\neg B]$?
There is the Law of Total Probability.
$$\newcommand{\P}{\operatorname{\sf P}}\begin{align}\P(A) & =\P(A\mid B)\P(B)+\P(A\mid B^\complement)\P(B^\complement) \end{align}$$
Which is about as concise an algebraic arrangement as you will get without involving quotients, such as:
$$\begin{align}\P(A\mid B)&=\dfrac{\P(A)-\P(A\mid B^\complement)\P(B^\complement)}{1-\P(B^\complement)}\end{align}$$