Say I have an exponentially distributed random variable $X$.
How does memorylessness of exponential distribution play a role in the conditional probability $P(\cdot \lvert X)$
For example, I have something along the lines of
$\mathbb P( X \leq t < X+ Y\lvert X)=\mathbb P( t < Y\lvert X)$ for another random variable $Y$.
I have been attempting to prove this, using the fact that by memorylessness of the exponential distribution, we have:
$$ \mathbb P (X-a \in \cdot \;\lvert X> a)=\mathbb P (X \in \cdot)\; \; \text{for any }a>0.$$
But I do not seem to get anywhere. Any ideas/help? Is independence of $X$ and $Y$ needed?
This is not true in general. The problem is that conditioning on the sigma algebra generated by $X$ is not the same as conditioning on the event that $X > t$, so the memoryless property does not apply here.
Let $F$ be the conditional cdf of $Y$ given $X$: $F(y,X) = P(Y\leq y|X)$ almost surely. For simplicity let's just assume that $Y$ conditioned on $X$ is a continuous random variable almost surely (that is, the conditional cdf of $Y$ given $X$ is almost surely continuous). Then,
$$P(X\leq t \leq X+Y|X) = \mathbb{I}_{X \leq t} P(Y \geq t - X|X) = \mathbb{I}_{X \leq t} (1 - F(t-X,X))\neq 1 - F(t,X) = P(t< Y|X).$$
Note, even if $Y$ is independent of $X$, then $F(y,X) := G(y)$ doesn't depend on $X$ so,
$$P(X\leq t\leq X+Y|X) = \mathbb{I}_{X\leq t}P(Y \geq t - X|X) = \mathbb{I}_{X\leq t}(1 - G(t- X)) \neq 1 - G(t) = P(t < Y).$$
It is possible to construct $Y$ for which your statement holds. For example, if $P(Y > t|X) = 1$ whenever $X \leq t$ and $P(Y > t|X) = 0$ whenever $X > t$, then the equation above will hold. However, it doesn't hold in any sort of generality.