This question is a great setup for explaining a better way of describing particles in quantum theory, one that bridges the traditional gap between single-particle quantum mechanics and quantum field theory (QFT).
I'll start with a little QFT, but don't let that scare you. It's easy, both conceptually and mathematically. In fact, it's easier than the traditional formulation of single-particle quantum-mechanics, both conceptually and mathematically! And it makes the question easy to answer, both conceptually and mathematically.
To make things easier, here's a little QFT
In QFT, observables are tied to space, not to particles. That's the most important thing to understand about QFT. Instead of assigning a "position observable" to each particle (which would be impossible in many models because particles can be created and destroyed), we assign detection observables to regions of space. Let $D(R)$ denote an detection observable associated with region $R$. In nonrelativistic QFT, the eigenvalues of $D(R)$ are natural numbers ($0$, $1$, $2$, ...) representing the number of particles found in $R$ when $D(R)$ is measured. If the model has more than one species of particle, then we have different detection observables for each species.
By the way, the traditional bit about "indistinguishable particles" is really just an obtuse way of saying what I said above: observables are tied to space, not to particles. Very simple.
Now, the difference between nonrelativistic QFT and nonrelativistic single-particle quantum mechanics is almost trivial: in the latter, the eigenvalues of $D(R)$ are restricted to $0$ and $1$.
As an example, consider ordinary nonrelativistic single-particle quantum mechanics, in one-dimensional space for simplicity. The traditional formulation uses a position operator $X$, whose measurement magically returns the particle's spatial coordinate. That formulation is convenient for some purposes, but it also causes trouble:
First, it causes conceptual trouble, because this is not how real-world measurements work: there is no measuring device that magically tells us the particle's coordinates no matter where it is in the universe. Real-world measurements are localized — they correspond to observables that are tied to regions of space, not tied to particles.
Second, it causes mathematical trouble, because $X$ has a continuous spectrum — it doesn't have any (normalizable) eigenstates, so it can't be perfectly measured even on paper.
Fortunately, we don't really need $X$. We can do better, both conceptually and mathematically, by using a collection of projection operators $D(R)$ instead. For any given spatial region $R$, the definition of $D(R)$ is simple:
If the particle's wavefunction is concentrated entirely within $R$, then it's an eigenstate of $D(R)$ with eigenvalue $1$.
If the particle's wavefunction is concentrated entirely outside of $R$, then it's an eigenstate of $D(R)$ with eigenvalue $0$.
In other words, $D(R)$ counts the number of particles in $R$, and since the model only has one particle, the answer can only be $0$ or $1$. If the particle's wavefunction is partly inside $R$ and partly outside $R$, then the result of the measurement will be $1$ with probability
$$
p = \frac{\langle\psi|D(R)|\psi\rangle}{\langle\psi|\psi\rangle},
\tag{1}
$$
and the result will be $0$ with the opposite probability $1-p$.
That's nicer conceptually, because it's closer to how we do things in the real world. It's also nicer mathematically, because $D(R)$ is a bounded operator with a discrete spectrum.
Note that the observables $D(R)$ for different regions $R$ all commute with each other (this should be obvious from the definition), so we can measure them all simultaneously if we want to — just like we can use an array of detectors in the real world.
The traditional position operator
How are the operators $D(R)$ related to the usual position operator $X$? Simple:
$$
X \approx \sum_n x_n D(R_n)
\tag{2}
$$
where the sum is over a set of regions $R_n$ that partition all of space into non-overlapping cells, and $x_n$ is the coordinate of some point in the $n$-th cell. In the limit as the size of the cells approaches zero, this approaches the usual position operator $X$. Conceptually, we're just filling space with an array of little detectors. We know where each one them is located, so if the $n$-th one detects the particle, then we know where the particle is.
To be fair, the traditional position operator (2) has some advantages over the detection observables $D(R)$. One advantage is that it allows us to use summary-statistics like averages (think of Ehrenfest's theorem) and standard deviations (think of the traditional uncertainty principles). This is just like the usual situation in statistics: the information is in the distribution (the results of measuring lots of $D(R)$s), but choosing a convenient labeling scheme lets us define things like averages and standard deviations to convey some incomplete but concise information about the distribution.
The particle on a circle
Now, the answer to the question should be obvious. Does a particle living on a circle have a position operator? Well, it certainly does have the operators $D(R)$, where now $R$ is any portion of the circle. Those operators are defined just like before. We can also define something analogous to $X$ if we really want to, like this:
$$
\Theta \approx \sum_n \theta_n D(R_n)
\tag{3}
$$
or like this:
$$
U \approx \sum_n e^{i\theta_n} D(R_n),
\tag{4}
$$
where the sum is over a partition of the circle into non-overlapping intervals $R_n$, and $\theta_n$ is an angular coordinate within the $n$-th interval. We can take the limit as the size of the intervals goes to zero, if we want to.
The operator (3) is most closely analogous to (2), and it's perfectly well-defined mathematically, but it's unnatural because the angular coordinate must have a discontinuity ($2\pi$ jump) somewhere on the circle. That makes the operator (3) less useful for talking about things like averages and standard deviations, but that's not quantum theory's fault. It's just mundane statistics: averages and standard deviations are most useful when using a monotonically increasing coordinate, which we can't do everywhere on a circle if we want the coordinate to be single-valued.
The operator (4) is more natural mathematically, because $e^{i\theta}$ is continuous everywhere around the circle. Beginners might complain that (4) is not hermitian, but observables do not need to be hermitian (see this question), contrary to what some introductions say. Unitary operators like (4) can also be used as observables, because quantum theory doesn't care about the coefficients of the projection operators, it only cares about the projection operators. The coefficients are useful for talking about averages and standard deviations, but we don't really need to talk about those things. They're convenient, but they're not necessary. Again, this is just like in ordinary statistics. Labels can be convenient, but they're not necessary.
So... does a particle on a circle have a position operator? It depends. What properties do you want the position operator to have? It does have the observables that really matter, which are the detection-observables $D(R)$. And we can use those to construct operators like (3) and (4) that each resemble the position operator (2) in different ways, but of course we can't have an operator that is completely like (2) because a circle is not completely like the real line. So, whether the answer is "yes" or "no" depends on exactly what you want, but at least the conceptual obstacles should be gone now.
