Consider the distribution of $X\vert X>Y$ or the memorylessness. Then see if it is easy to change it to $X-Y\vert X>Y$
But here Y is also a random variable. Does it matter?
If you know $\mathbb{P}(X \vert Y)$$\mathbb{P}(A \vert B)$ and $Y$$B$ is itself a random variable, then you can find the probability of $\mathbb{P}(X)$$\mathbb{P}(A)$ as a compound distribution or by using the law of total probability
$$\mathbb{P}(X) = \sum_{\forall Y} \mathbb{P}(X \vert Y)\mathbb{P}(Y) \underbrace{ = \mathbb{P}(X \vert Y) \sum_{\forall Y} \mathbb{P}(Y) = \mathbb{P}(X \vert Y)}_{\text{if $ \mathbb{P}(X \vert Y)$ is independent of $Y$}} $$$$\mathbb{P}(A) = \sum_{\forall B} \mathbb{P}(A \vert B)\mathbb{P}(B) $$
if $ \mathbb{P}(A \vert B) = f(A)$ is a function independent of $B$ then it can be taken out of the sum and you get
$$\mathbb{P}(A) = \sum_{\forall B} \mathbb{P}(A \vert B)\mathbb{P}(B) = f(A)\sum_{\forall B} \mathbb{P}(B) = f(A) $$
Similarly when $\mathbb{P}(X-Y\vert X>Y, Y)$ is independent from $Y$ then you know $\mathbb{P}(X-Y\vert X>Y)$