I'll assume that the Earth is a sphere of radius 1 and that the starting point is in the Northern hemisphere.
In the accompanying Wikipedia article about the rhumb line, it tells us that "The distance between two points, measured along a [rhumb line], is simply the absolute value of the secant of the bearing (azimuth) times the north-south distance". Hence, if our bearing returns us to the starting point, it will have the correct distance as specified in the question (since $\sec(\pi/4) = \sqrt{2}$). So, it suffices to find the latitudes for which the north-easterly journey returns us to the beginning. Now, if we travel south from a latitude of $\phi_1$ to a distance $\pi/2$ and then West for the same distance, the azimuthal angle we go through is $\frac{\pi}{2\sin(\phi_1)}$.
Another part of the Wikipedia article gives us that
The differential relationship between the azimuthal angle $\lambda$ and the polar angle $\phi$ as we travel along the rhumb line is $\frac{d\lambda}{d\phi} = \sec \phi$. Integrating both sides with respect to $\phi$ tells us that travelling along the rhumb from polar angle $\phi_0$ to $\phi_1$ corresponds to travelling an azimuth of angle $\log(\tan(\phi_1) + \sec(\phi_1)) - \log(\tan(\phi_0) + \sec(\phi_0))$. Now, as per the question, we want $\phi_0 = \phi_1 - \pi/2$, hence the azimuth difference in the last leg of the journey is $\log(\frac{\tan(\phi_1) + \sec(\phi_1)}{-\cot(\phi_1) + \csc(\phi_1)})$ which must be equal to the azimuthal angle $\frac{\pi}{2\sin(\phi_1)}$ we derived from travelling south then west.
So any latitude which works is the solution $x$ to
$\log(\frac{\tan(x) + \sec(x)}{-\cot(x) + \csc(x)}) = \frac{\pi}{2\sin(x)}$ Plotting them both, it seems like there is one solution at latitude $x \approx 1.019$ radians or a latitude of $58.4$ degrees, roughly.
However, there's a twist in the tale because, of course, as @human pointed out,
The above equation is only valid modulo $2\pi$ so we can add an integer multiple of $2\pi$ to the left hand side of the equation and also get a valid solution i.e, $\log(\frac{\tan(x) + \sec(x)}{-\cot(x) + \csc(x)}) + 2k\pi = \frac{\pi}{2\sin(x)}$ for $ k \in \mathbb{Z}$. The right hand side is strictly decreasing on the interval $(0, \pi/2)$ and the left hand side is convex with limits of $+\infty$ at both ends of the interval and a steeper slope at zero so for each value of $k$ there is one intersection point between the curves i.e, one solution.
For negative $k$
The last leg of the journey "encircles the Earth" (in the sense of intersecting the same line of longitude) $k$ more times than the second leg. The solutions are always very close to $\pi/2$ i.e, the North Pole, even when $k=-1$. This is a result of the fact that the rhumb line spins infinitely many times as it approaches the North Pole.
For positive $k$
The second leg "encircles the Earth" $k$ more times than the last leg. These solutions tend more gradually towards (zero) the equator as $k$ increases.
What about the limits?
As $k \rightarrow \infty$, the solution goes to the equator. At the equator, a $\pi/2$ movement south brings us to the South Pole from where all directions (except North) are ill-defined so we don't have a solution here. As $k \rightarrow -\infty$, however, the solution tends to the North Pole. The North Pole itself constitutes a valid and very interesting solution. After the first two legs of the journey we are at the equator. From there, travelling along the rhumb line, we wind around the North Pole an infinite number of times in order to reach it but do so in a finite distance ($\sqrt{2} \frac{\pi}{2}$). This is one of the incredible facts about the rhumb line. It is seemingly infinite at both ends but is finite in length (rather counterintuitive).