I am trying to understand the relationship among the following: $P(A | B)$, $P(A | \lnot B)$, and $P(A)$
A is independent of B iff: $P(A | B) = P(A)$
But, if A is independent of B, then does this necessarily imply that: $P(A | B) = P(A | \lnot B)$
And, if the A and B are not independent, does that necessarily imply that: $P(A | B) \neq P(A | \lnot B)$
If the above all does hold, does it fail to always hold as soon as we add another outcome possibility to either random variable (i.e., A or B is not dichotomous).
Yes, if $A$ and $B$ are independent then $P(A|B) = P(A|\lnot B)$. To see this, write $P(A)$ by conditioning on $B$: $$P(A) = P(A|B)P(B) + P(A|\lnot B)P(\lnot B) = P(A|B)P(B) + P(A|\lnot B)(1-P(B))$$ Since $A$ is independent of $B$, it is independent of $\lnot B$ (exercise) so it follows that $P(A|\lnot B) = P(A)$ and we have $$0 = P(A|B)P(B)- P(A|\lnot B)P(B)$$ Thus $P(A|B) = P(A|\lnot B)$.
When $A$ and $B$ are not independent then we must have $P(A|B) \neq P(A|\lnot B)$. Notice that this is exactly the converse of the previous statement, so we can simply show that $P(A|B) = P(A|\lnot B)$ implies that $A$ and $B$ are independent. $$\frac{P(A\cap B)}{P(B)} = \frac{P(A\cap \lnot B)}{P(\lnot B)} = \frac{P(A) - P(A\cap B)}{1 - P(B)}\implies \frac{P(A\cap B)}{P(B)} - P(A\cap B) + P(A\cap B) = P(A).$$ Thus $P(A|B) = P(A)$ and $A$ and $B$ are independent.
I am not sure what you are asking about in the in last part. A random variable can take a range of values, but the probabilities in-terms of an event i.e. the event $X = x$ or the random variable $X$ takes on a certain value $x$. An event is either true or false so the above analysis still holds.