To find the probability density you need to have an intuitive understanding of the motion under the force. For the harmonic oscillator, the particle is always oscillating from $x=-A$ to $x=+A$. Each cycle is identical to the previous one, and so the probability of finding the particle between $x$ and $x+dx$ is $dt/T$ where dt is the time the particle takes to move from $x$ to $x+dx$ and $T$ is the total time period of one oscillation. Since the particle moves the slowest near the amplitudes, it will spend a larger fraction of the total period there compared to other points, and so you would expect the probability density to be highest at those points. Which is the case, as you can see from the equation you wrote. So you were right on that point - the classical probability density should be proportional to $1/v$.
Now, in your case the potential is $k/x$, which is an attractive potential towards the center. The force gets infinitely large as you approach the center. So there is no oscillation here. So your anticipation that there is "some well characterized periodic, oscillating motion with a period defined in terms of $k$ and $m$" was wrong. The particle starts with some initial velocity $v_0$ at some initial position $x_0$. If the initial velocity is towards the center, then it approaches the center and then stops there. If the initial velocity is away from the center, then it moves away at first and then turns and approaches the center and stops there.
Nevertheless, it is possible to find the probability density if you properly define the problem. You would expect the probability density to be a Dirac delta function because the particle spends an infinite amount of time at the center and a negligible amount of time (compared to infinity) at other points. However, if you want the probability density of finding the particle during the course of it's trajectory from $x_0$ to $x=0$, then first find the time it takes to go there as a function of $x$,
$$T = \int dt = \int_{x_0}^0 \frac{1}{v}dx$$
$$ = \lim_{x \to 0} \int_{x_0}^x \pm \frac{1}{\sqrt{(2/m)(E - V(x))}}$$
where I used $E = mv^2/2 + V(x)$. The value of $E$ is given by the initial conditions
$$E = \frac{1}{2}mv_0^2 + V(x_0)$$
The limit sign is there because I don't know what the integral comes out to be, but since the potential goes to infinity as $x \to 0$, just putting $0$ in the upper limit might give absurd results. The plus minus sign is related to the initial condition. If the initial velocity is to the right, then you choose the plus sign. If on top of that the initial position is at positive $x$, then you break the integral into two parts. You choose the plus sign for the integral from $x_0$ to $x^*$, the turning point. And then you choose the minus sign for the integral from $x^*$ to $0$, and add up both the integrals. Similar considerations apply when the initial velocity is to the left.
Now after finding $T$, the rest is easy. $\rho dx$, which is the probability of finding the particle between $x$ and $x+dx$ is also $dt/T$. And so
$$\rho dx = \frac{dt}{T}$$
$$\rho = \frac{1}{T} \frac{1}{|dx/dt|} = \frac{1}{|v|T}$$
$$\rho(x) = \frac{1}{T \sqrt{(2/m)(E - V(x))}}$$