Probability that $x$ divides $k \leq n$ and an equation question. The exact probability that a fixed positive integer $x \leq n$ divides a randomly selected positive integer $k \leq n$ is $\dfrac{\lfloor \dfrac{n}{x} \rfloor}{n}$ which is the same as $\dfrac{n - n_{(x)}}{xn}$ where $n_{(x)}$ is defined to be the smallest positive residue mod $x$. 
So the probability that a randomly selected positive integer $k \leq n$ is divisible by $x$ or $y$ is $P(x, y) = \dfrac{n - n_{(x)}}{xn} + \dfrac{n - n_{(y)}}{yn} - \dfrac{n - n_{(xy)}}{xyn}$.
Setting this equal to $1$ gives $y(n - n_{(x)}) + x(n - n_{(y)}) - (n - n_{(xy)}) = xyn = n(y + x - 1) - yn_{(x)} - xn_{(y)} + n_{(xy)}$
So is $x = 2, y = 3, n = 4$ the unique solution to this weird equation?

*

*Your expression for $P(x, y)$ does not hold generally. To be more precise, it should be
$$
P(x, y) = P(x) + P(y) - P(\mathsf{lcm}(x,y))
$$
where $P(x) = \left(n - n_{(x)}\right)/{nx}$, and $\mathsf{lcm}(x, y)$ is the least common multiple of $x$ and $y$.

*Any combination with $x = 1$, $1 \leq y \leq n$ is a solution to the equation since
$$
P(1, y) = P(1) + P(y) - P(\mathsf{lcm}(x, y)) = P(1) + P(y) - P(y) = P(1) = 1
$$
