In a conditional probability distribution $P(A|B)$, can I substitute a new random variable $C$ to represent "A given B" and write it as $P(C) = P(A|B)$?
If so, would this expression be true? $$P(C|D) = P(A|B,D)$$
Please forgive my ignorance and correct me if I'm wrong, I've taken courses on probability but don't know where the gaps in my knowledge lie.
I see nothing wrong with this. For example the notation $P(A|B=b)$ means there is a random variable, chosen to be denoted as $A|B=b$, and whose probability we write as $P(A|B=b)$. You can certainly replace $A|B=b$ with anything you like. Using your $C$ there is no restriction on then asking what is the distribution of $C|D=d$, i.e we want $P(C|D=d)$. Omitting the constants for clarity;
\begin{align} P(C|D)&=P(A|B,D)\\ \Longleftrightarrow\frac{P(C,D)}{P(D)}&=P(A|B,D)\\ \Longleftrightarrow\frac{P(A|B,D)}{P(D)}&=P(A|B,D), \end{align}
which holds if $P(D)=1$, i.e. if $D$ is a constant and not a random variable. Thus in general your notation of "carrying D along" with B after the "|" notation will only be correct for constants.
You see this in non-time-series regression contexts a lot, for example you might see $Y|U=u\sim F(\beta_{0},\beta_{1},X)$ to mean a conditional distribution induced by the model $Y|\{U=u\}=\beta_{0}+\beta_{1}X+\epsilon$ for $\epsilon$ random, and where $U\sim N(0,\sigma^{2}_{u})$ is a random coefficient or "random effect". We can write $P(Y|U,\beta_{0},\beta_{1},X)$ for our probability, and this is equal to $P[C=\{Y|U\}|D=\{\beta_{0},\beta_{1},X\}]$ because $D$ is constant. For other regression models in time-series settings the covariate set $X$ is assumed random, and then the distribution $P(\{Y|U\}|X)$ is not neccessarily equal to $P(Y|U,X)$.
It is a typical mistake of confused freshmen to think that $\Pr(A\mid B)$ is the probability of some object called $\text{“}A\text{ given }B\text{''}$ or $\text{“}A\mid B\text{''}.$ It is in fact the probability-given-$B$, of $A.$
However, you could let $C = A\cap B$ and then in a new sample space consisting only of $B,$ the conditional probability $\Pr(A\mid B)$ would become $\Pr(C).$
