I think this is very similar to what tkf suggested! See theorem 2.2 of
https://www.uni-muenster.de/imperia/md/content/theoretische_mathematik/diffgeo/mr1783960.pdf
Suppose the compact group G acts transitively on M. Then the universal cover
$$
\widetilde{G}\cong K \times \mathbb{R}^n
$$
also acts transitively (here $ K $ is compact and simply connected). So for some closed stabilizer group $ H $ we have
$$
M \cong \widetilde{G}/H
$$
Consider the projection p of H onto the second factor $ \mathbb{R}^n $. Then we have a SES
$$
1 \to ker(p) \to H \to p(H) \to 1
$$
Observe that p(H) is abelian since it is a subgroup of $ \mathbb{R}^n $ and $ ker(p) $ is compact because it is a subgroup of $ K \times 0 $. Take $ \pi_0 $ of this SES then we have another SES of discrete groups
$$
1\to \pi_0(ker(p)) \to \pi_0(H) \to \pi_0(p(H)) \to 1
$$
The first component group is compact and discrete so it must be finite. The third component group is a discrete subgroup $ \mathbb{R}^n $ so just $ \mathbb{Z}^k $. This proves that $\pi_0(H) $ is finite by abelian. Since $ \widetilde{G} $ is simply connected then $ \pi_1(M)=\pi_0(H) $ is finite by abelian.
Related fun facts from this same reference:
If G compact acts transitively then M admits a nonnegative curvature metric with respect to which G acts by isometries.
Let $F$ be a path-connected topological space and $\phi : F \to F$ a homeomorphism. Denote the mapping torus of $\phi$ by $M$ so that we have a fiber bundle $F \to M \to S^1$. Explicitly, $M = F\times[0, 1]/\sim$ where $(x, 0) \sim (\phi(x), 1)$. From the long exact sequence in homotopy, there is a short exact sequence $$0 \to \pi_1(F) \to \pi_1(M) \to \pi_1(S^1) \to 0.$$ The covering space of $M$ corresponding to the image of $\pi_1(F)$ in $\pi_1(M)$ is homeomorphic to $F\times\mathbb{R}$. This follows from the fact that $M$ can be constructed as the quotient of $F\times\mathbb{R}$ by the $\mathbb{Z}$-action generated by $(x, t) \mapsto (\phi(x), t + 1)$. In particular, the universal cover of $M$ is homeomorphic to $\widetilde{F}\times\mathbb{R}$ where $\widetilde{F}$ is the universal cover of $F$.
Suppose now that $F$ is a closed $(n-1)$-manifold, so that $M$ is a closed $n$-manifold. If $\pi_1(M) \cong \mathbb{Z}$, then $\pi_1(F) = 0$ by the short exact sequence above. As suggested by Moishe Kohan, let $N = M \# X$ where $X$ is a closed simply connected $n$-manifold, then $\widetilde{N}$, the universal cover of $N$, is homeomorphic to the connected sum of $\widetilde{F}\times\mathbb{R} = F\times\mathbb{R}$ and countably many copies of $X$. If $H_k(X; \mathbb{Z}) \neq 0$ for some $0 < k < n$, then it follows from Mayer-Vietoris that $H_k(\widetilde{N}; \mathbb{Z}) \cong H_k(\widetilde{F}; \mathbb{Z})\oplus\bigoplus_{i\in\mathbb{Z}}H_k(X)$; in particular, $H_k(\widetilde{N}; \mathbb{Z})$ is not finitely generated. As $\pi_1(N) \cong \pi_1(M) \cong \mathbb{Z}$, if $N$ were a mapping torus, the fiber $F'$ would be a closed simply connected manifold, so $\widetilde{N}$ would be homeomorphic to $\widetilde{F'}\times\mathbb{R} = F'\times\mathbb{R}$ which has finitely generated homology. Therefore $N$ is not a mapping torus. Note, if $H_k(X; \mathbb{Z}) = 0$ for all $0 < k < n$, then $X$ is a simply connected homology sphere and hence homeomorphic to $S^n$, in which case $N$ is homeomorphic to $M$ (which is a mapping torus).
I believe that the hypothesis $\pi_1(M) \cong \mathbb{Z}$ is unnecessary, but the proof above only extends to the case where $\pi_1(M)$ is an extension of $\mathbb{Z}$ by a finite group, i.e. $F$ has finite fundamental group.
Best Answer
Because you're assuming the group is compact, there's another method for classifying such spaces. This approach can be used in higher dimensions (though people often assume the resulting manifold is simply connected just to cut down on the work).
I'll also always be assuming things are connected: the components of a homogeneous space $G/H$ are always diffeomorphic, and the identity component of $G$ acts transitively on a connected component of $G/H$, so this is all ok.
Suppose $G$ is a connected compact Lie group, $H\subseteq G$ is a closed subgroup. Because $G$ is compact, we may equip the homogeneous space $G/H$ with a Riemannian metric for which the action by $G$ is isometric.
Now, consider the isotropy action of $H$ on $T_{eG} G/H$, where $e\in G$ denotes the identity element. Recall (see, e.g., this MSE question) that if an isometry of a (connected) Riemannian manifold fixes a point and acts as the identity at the tangent space at that point, then that isometry must be the identity. Elements of $h$ fix $e H\in G/H$, so, if such an $h$ acts trivially on $T_{eG}G/H$, we must have $h = e$.
Said another way, the isotropy action gives an injective map $H\rightarrow O(T_{eG} G/H)$ into the orthogonal group. Thus, we may view $H$ as a subgroup of $O(T_{eG} G/H)$.
I'll now specialize to the case where $G/H$ is $3$-dimensional, so $O(T_{eG} G/H)$ can be identified with $O(3)$. The connected subgroups of $SO(3)$ are well known: they are $\{e\}, SO(3)$, or a conjugate of the usual $SO(2)\subseteq SO(3)$. Note that $G/H$ is equivariantly diffeomorphic to $G/(gHg^{-1})$ for any $g\in G$, so, in terms of classification, we can assume the identity component $H^0$ of $H$ is given by one of $\{e\}, SO(2)$, or $SO(3)$.
We'll break into cases depending on $H^0$.
Case A
If $H^0 = \{e\}$, then $3 = \dim G - \dim H$ impies that $\dim G = 3$. From the classification of simply connected Lie groups, $G$ has a cover which is either isomorphic to $SU(2)$ or to $T^3$. In the first case $G = SU(2)$ or $SO(3)$ and $SO(3)/H = SU(2)/\pi^{-1}(H)$ with $\pi:SU(2)\rightarrow SO(3)$ the double cover, so you just get quotients of $SU(2)$. In the second case, you get quotients of $T^3$. However, $H\subseteq T^3$ is normal, so $T^3/H$ is an abelian Lie group of dimension $3$, so $T^3/H\cong T^3$ no matter what $H$ is (so long as $H^0 = \{e\}$.)
Case B
If $H^0 = SO(2)$, then $\dim G = 4$. From the classification of simply connected Lie groups, $G$ has a cover of the form $G = T^4$ or $G = SU(2)\times S^1$. If $G = T^4$, the argument from the previous paragraph establishes that $T^4/H\cong T^3$. So, we will assume that $G$ is covered by $G = SU(2)\times S^1$. In fact, we'll pull everything back: $G/H\cong (SU(2)\times S^1)/\pi^{-1}(H)$, so we'll actually work with $G= SU(2)times S^1$. (Note though, that by pulling back, $\pi^{-1}(H)$ need not act effectively on $(SU(2)\times S^1)/\pi^{-1}(H)$) any more.) Also, instead of writing $\pi^{-1}(H)$ everywhere, I'll abuse notation and just write $H$.
Lemma: Suppose $H$ acts diagonally on $G_1\times G_2$ and that the action of $H$ on $G_2$ is transitive with isotropy group $H'$. Then $(G_1\times G_2)/H$ is canonically diffeomorphic to $G_1/H'$.
Proof: Just check that the map $G_1/H'\rightarrow (G_1\times G_2)/H$ sending $g_1 H'$ to $(g_1,e)H$ is a diffeomorphism. $\square$
Using the lemma, it follows that if the projection of $H^0$ to the $S^1$ factor of $G$ is surjective, then we can write $G/H\cong SU(2)/H'$, where $H'$ is $0$-dimensional. These were classified in Case A). So, we may assume that the projection of $H^0$ to the $S^1$ factor of $G$ is trivial. That is, we may assume $H^0$ acts only on the $SU(2)$ factor of $G$, so $G/H^0\cong S^2\times S^1$.
So, we understand $H^0$ in this case, but what about $H$? Well, let's assume $H\neq H^0$, and pick $(h_1,h_2)\in H\setminus H_0$. Because the normalizer $N:=N_{SU(2)}(SO(2))$ has two components, and $h_1 \in N$, we know that $(h_1,h_2)^2$ acts as a rotation on just the $S^1$ factor of $G/H^0 \cong S^2\times S^1$. Hence, $(G/H^0)/\langle (h_1,h_2)^2\rangle \cong S^2\times S^1$, so we may as well assume that $(h_1,h_2)^2 \in H^0$, so that, in particular, $h_2 = \pm 1$. That is, we may as well assume that $H$ has precisely two components. If $h_1\in SO(2)$, then $G/H$ is diffeomorphic to $S^2\times S^1$ independent of $h_2$. If $h_1\notin SO(2)$ and $h_2 = 1$, we get $\mathbb{R}P^2\times S^1$, and if $h_1\notin SO(2)$ and $h_2 = -1$, we get the non-trivial $S^2$-bundle over $S^1$ (the quotient of $S^2\times S^1$ by the diagonal antipodal action).
This concludes case B.
Case C In this case, $H^0 = SO(3)$. Then $\dim G = 6$, so $G$ is covered by one of $T^6, SU(2)\times T^3$, or $SU(2)\times SU(2)$.
For $G=T^6$, because $H^0$ is non-abelian, there is no $H^0\subseteq G$, even up to cover.
For the other two, by replacing $H^0$ with a cover of $H^0$ (i.e., $SU(2)$), we can actually assume $G= SU(2)\times T^3$ or $G = SU(2)\times SU(2)$. However, in both cases, it is easy to see that the above Lemma applied, so we find $G/H\cong T^3/H'$ or $SU(2)/H'$ where $H'$ is $0$-dimensional, getting us back to Case A. This completes Case C.