I'll put my comments together and post them as an answer just for clarity. I expect you won't be very satisfied, however, as all I do is reduce your problem to another problem.

My notation:

$Q$ is a RV whose support is $\mathbb{Q}\cap\left[0,1\right]$ -- my $Q$ is not the same as the $Q$ the OP constructs from his $\frac{X}{Y}$. We'll define this $Q$ using $Y$ and $f$, which I introduce below.

$Y$ is any RV whose support is $\mathbb{N}\equiv\left\{1, 2, \ldots\right\}$ -- the $Y$ given by the OP would work, for example.

$f$ is any one-to-one correspondence $f:\mathbb{N}\rightarrow\mathbb{Q}\cap\left[0,1\right]$ and $f^{-1}$ is its inverse. We know these exist.

Now I claim I can reduce your problem to just finding an $f$ and its $f^{-1}$:

Just let $Q=f^{-1}\left(Y\right)$ and you are done. The PMF of $Q$ is $\Pr[Q =q] = \Pr[Y = f^{-1}(q)]$.

I'll put my comments together and post them as an answer just for clarity. I expect you won't be very satisfied, however, as all I do is reduce your problem to another problem.

My notation:

$Q$ is a RV whose support is $\mathbb{Q}\cap\left[0,1\right]$ -- my $Q$ is not the same as the $Q$ the OP constructs from his $\frac{X}{Y}$. We'll define this $Q$ using $Y$ and $f$, which I introduce below.

$Y$ is any RV whose support is $\mathbb{N}\equiv\left\{1, 2, \ldots\right\}$ -- the $Y$ given by the OP would work, for example.

$f$ is any one-to-one correspondence $f:\mathbb{N}\rightarrow\mathbb{Q}\cap\left[0,1\right]$ and $f^{-1}$ is its inverse. We know these exist.

Now I claim I can reduce your problem to just finding an $f$ and its $f^{-1}$:

Just let $Q=f\left(Y\right)$ and you are done. The PMF of $Q$ is $\Pr[Q =q] = \Pr[Y = f^{-1}(q)]$.

Edit:

Here is a function g that plays the role of $f$, despite not being a one-to-one correspondence (because of duplicates):

g <- function(y) {
    y <- as.integer(y)
    stopifnot(y >= 1)
    b <- 0
    a <- 0
    for (unused_index in seq(1, y)) {
        if (a >= b) {
            b <- b+1
            a <- 0
        } else {
            a <- a+1
        }
    }
    return(sprintf("q = %s / %s", a, b))
    ## return(a / b)
}

I'll put my comments together and post them as an answer just for clarity. I expect you won't be very satisfied, however, as all I do is reduce your problem to another problem.

My notation:

$Q$ is a RV whose support is $\mathbb{Q}\cap\left[0,1\right]$ -- my $Q$ is not the same as the $Q$ the OP constructs from his $\frac{X}{Y}$.

$Y$ is any RV whose support is $\mathbb{N}\equiv\left\{1, 2, \ldots\right\}$ -- the $Y$ given by the OP would work, for example.

$f$ is any one-to-one correspondence $f:\mathbb{N}\rightarrow\mathbb{Q}\cap\left[0,1\right]$ and $f^{-1}$ is its inverse. We know these exist.

Now I claim I can reduce your problem to just finding an $f$ and its $f^{-1}$:

Just let $Q=f^{-1}\left(Y\right)$ and you are done. The PMF of $Q$ is $\Pr[Q =q] = \Pr[Y = f^{-1}(q)]$.

I'll put my comments together and post them as an answer just for clarity. I expect you won't be very satisfied, however, as all I do is reduce your problem to another problem.

My notation:

$Q$ is a RV whose support is $\mathbb{Q}\cap\left[0,1\right]$ -- my $Q$ is not the same as the $Q$ the OP constructs from his $\frac{X}{Y}$. We'll define this $Q$ using $Y$ and $f$, which I introduce below.

$Y$ is any RV whose support is $\mathbb{N}\equiv\left\{1, 2, \ldots\right\}$ -- the $Y$ given by the OP would work, for example.

$f$ is any one-to-one correspondence $f:\mathbb{N}\rightarrow\mathbb{Q}\cap\left[0,1\right]$ and $f^{-1}$ is its inverse. We know these exist.

Now I claim I can reduce your problem to just finding an $f$ and its $f^{-1}$:

Just let $Q=f^{-1}\left(Y\right)$ and you are done. The PMF of $Q$ is $\Pr[Q =q] = \Pr[Y = f^{-1}(q)]$.