The claim is **true**. (As proved here, it is quite unique to sinusoidal vector fields. The same reference also mentions, in its introduction, existing applications of equation (1) with $g$ taking the particular form mentioned above.)

In short (as in Theorem 1 of here): the stereographic projections of the solutions of (1) satisfy a Riccati equation (1st order quadratic ODE), and (as in Proposition 2 of here) the time-$t$ mappings of Riccati equations are Möbius transformations; the classification of Möbius transformations then gives the result. In fact, **either: [A] all solutions of (1) are neutrally stable; or [B] (1) has a unique asymptotically stable $1$-periodic solution and a unique unstable $1$-periodic solution, and all solutions other than the unstable $1$-periodic solution are attracted to the stable $1$-periodic solution; or [C] (1) has a unique $1$-periodic solution, and this solution attracts all solutions but is unstable.**

To go through the proof in full detail:

**Lemma 1.** For any solution $\theta(\cdot)$ of (1), letting $y(\cdot)$ be the stereographic projection $y(t)=\tan(\frac{\theta(t)}{2})$, we have that
$$ \dot{y}(t) \ = \ Ay(t) \ + \ \tfrac{1}{2}g(t)(1+y(t)^2) $$
whenever $\theta(t) \neq \pi$.

*Proof.* Straightforward direct computation. $\ \square$

Now let $\mathbb{S}_4$ be the set of points in $(\mathbb{S}^1)^4$ whose four coordinates are distinct. Define $C \colon \mathbb{S}_4 \to \mathbb{R}$ to be the cross-ratio of the stereographic projections $\tan(\frac{\cdot}{2})$ of the four inputs; note that $C$ is a smooth function, as it can be written explicitly (i.e. without reference to arithmetic with infinity) as
$$ C(\theta_1,\theta_2,\theta_3,\theta_4) \ = \ \left\{ \begin{array}{c l} \frac{\left(\tan(\frac{\theta_3}{2}) \tan(\frac{\pi-\theta_1}{2})-1\right)\left(\tan(\frac{\theta_4}{2}) - \tan(\frac{\theta_2}{2})\right)}{\left(\tan(\frac{\theta_3}{2}) - \tan(\frac{\theta_2}{2})\right)\left(\tan(\frac{\theta_4}{2}) \tan(\frac{\pi-\theta_1}{2})-1\right)} & \theta_1 \neq 0 \textrm{ and } \theta_2,\theta_3,\theta_4 \neq \pi \\ & \\ \frac{\left(\tan(\frac{\theta_3}{2}) - \tan(\frac{\theta_1}{2})\right)\left(\tan(\frac{\theta_4}{2}) \tan(\frac{\pi-\theta_2}{2})-1\right)}{\left(\tan(\frac{\theta_3}{2}) \tan(\frac{\pi-\theta_2}{2})-1\right)\left(\tan(\frac{\theta_4}{2}) - \tan(\frac{\theta_1}{2})\right)} & \theta_2 \neq 0 \textrm{ and } \theta_1,\theta_3,\theta_4 \neq \pi \\ & \\ \frac{\left(1-\tan(\frac{\pi-\theta_3}{2}) \tan(\frac{\theta_1}{2})\right)\left(\tan(\frac{\theta_4}{2}) - \tan(\frac{\theta_2}{2})\right)}{\left(1-\tan(\frac{\pi-\theta_3}{2}) \tan(\frac{\theta_2}{2})\right)\left(\tan(\frac{\theta_4}{2}) - \tan(\frac{\theta_1}{2})\right)} & \theta_3 \neq 0 \textrm{ and } \theta_1,\theta_2,\theta_4 \neq \pi \\ & \\ \frac{\left(\tan(\frac{\theta_3}{2}) - \tan(\frac{\theta_1}{2})\right)\left(1-\tan(\frac{\pi-\theta_4}{2}) \tan(\frac{\theta_2}{2})\right)}{\left(\tan(\frac{\theta_3}{2}) - \tan(\frac{\theta_2}{2})\right)\left(1-\tan(\frac{\pi-\theta_4}{2}) \tan(\frac{\theta_1}{2})\right)} & \theta_4 \neq 0 \textrm{ and } \theta_2,\theta_3,\theta_4 \neq \pi. \end{array} \right. $$

**Lemma 2.** $C(\cdot)$ is a conserved quantity for the four-point motion of (1).

For this, we need the following very simple fact:

**Lemma 3.** (A) For any linear map $L \colon \mathbb{R} \to \mathbb{R}$ and quantities $y_1,y_2,y_3 \in \mathbb{R}$,
$$ (y_2-y_1)(L(y_3)-L(y_1)) \ = \ (L(y_2)-L(y_1))(y_3-y_1). $$
(B) For any quadratic map $Q \colon \mathbb{R} \to \mathbb{R}$ and quantities $y_1,y_2,y_3,y_4 \in \mathbb{R}$,
\begin{align*}
& (y_3-y_1)(y_4-y_2)[(y_3-y_2)(Q(y_4)-Q(y_1)) + (Q(y_3)-Q(y_2))(y_4-y_1)] \\
= \ & [(y_3-y_1)(Q(y_4)-Q(y_2)) + (Q(y_3)-Q(y_1))(y_4-y_2)](y_3-y_2)(y_4-y_1).
\end{align*}

*Proof.* (A) Writing $L(x)=ax$, we have that
$$ \mathrm{LHS} \ = \ a(y_2-y_1)(y_3-y_1) \ = \ \mathrm{RHS}. $$
(B) Writing $Q(x)=ax^2+bx+c$, we have that
\begin{align*}
\mathrm{LHS} \ &= \ 2b(y_3-y_1)(y_4-y_2)(y_3-y_2)(y_4-y_1) \\
& \hspace{10mm} + a(y_3-y_1)(y_4-y_2)(y_3-y_2)(y_4^2-y_1^2) \\
& \hspace{10mm} + a(y_3-y_1)(y_4-y_2)(y_4-y_1)(y_3^2-y_2^2) \\
&= \ (y_3-y_1)(y_4-y_2)(y_3-y_2)(y_4-y_1)[2b + a(y_1+y_2+y_3+y_4)].
\end{align*}
One obtains the same for the $\mathrm{RHS}$. $\ \square$

*Proof of Lemma 2.* Let $\theta_1(t),\theta_2(t),\theta_3(t),\theta_4(t)$ be four distinct solutions of (1), with projections $y_1(t),y_2(t),y_3(t),y_4(t)$ respectively, and let $C_0(t)=C(\theta_1(t),\theta_2(t),\theta_3(t),\theta_4(t))$. We need that for all $t$, $\dot{C}_0(t)=0$. Firstly, if $U \subset \mathbb{R}$ is an open time-interval during which $\theta_i(t)$ is fixed at $\pi$ for some $i$, then $g(t)=0$ on $U$ (since $\sin(\pi)=0$) and
$$ C_0(t) \ = \ \pm\frac{y_j(t)-y_k(t)}{y_j(t)-y_l(t)} $$
on $U$; so one can use Lemma 1 (with $g(t)=0$) to compute $\dot{C}_0(t)$, and due to Lemma 3(A) one obtains that $\dot{C}_0(t)=0$. Secondly, if we have a time $t$ at which $\theta_1(t),\theta_2(t),\theta_3(t),\theta_4(t) \neq \pi$, then
$$ C_0(t) \ = \ \frac{(y_3(t)-y_1(t))(y_4(t)-y_2(t))}{(y_3(t)-y_2(t))(y_4(t)-y_1(t))} \, ; $$
so one can use Lemma 1 to compute $\dot{C}_0(t)$, and due to Lemma 3(B) one obtains that $\dot{C}_0(t)=0$. Now the solutions $\theta_i(t)$ are continuously differentiable (since $g$ is continuous), and so since $C$ is smooth, we have that $C_0(\cdot)$ is continuously differentiable. Thus we can conclude overall that $\dot{C}_0(t)=0$ for all $t$. $\ \square$

Now let $\hat{\mathbb{C}}=\mathbb{C} \cup \{\infty\}$ be the one-point compactification of $\mathbb{C}$, and let $\hat{\mathbb{R}}=\mathbb{R} \cup \{\infty\} \subset \hat{\mathbb{C}}$. A *Möbius transformation* is a map $T \colon \hat{\mathbb{C}} \to \hat{\mathbb{C}}$ taking either the form $T(z)=az+b$ with $a \neq 0$ or the form $T(z)=\frac{a}{z+b}+c$ with $a \neq 0$.

**Lemma 4.** Fix $t \geq 0$, and let $f \colon \mathbb{S}^1 \to \mathbb{S}^1$ be the time-$t$ mapping of (1). Let $F \colon \hat{\mathbb{R}} \to \hat{\mathbb{R}}$ be the conjugation of $f$ by the projection $\tan(\frac{\cdot}{2})$. Then $F$ is the restriction to $\hat{\mathbb{R}}$ of a Möbius transformation.

*Proof.* Pick any distinct points $y_1,y_2,y_3 \in \mathbb{R} \cap F^{-1}(\mathbb{R})$. For every $x \in (\mathbb{R} \setminus \{y_1,y_2,y_3\} ) \cap F^{-1}(\mathbb{R})$, Lemma 2 gives that
$$ \frac{(F(y_3)-F(y_1))(F(x)-F(y_2))}{(F(y_3)-F(y_2))(F(x)-F(y_1))} \ = \ \frac{(y_3-y_1)(x-y_2)}{(y_3-y_2)(x-y_1)}. $$
It is easy to see that one can rearrange the above equation to make $F(x)$ the subject, with the right-hand side being expressible as the ratio of two affine functions of $x$. Hence, since $F$ is continuous and injective, $F$ is a Möbius transformation (restricted to $\hat{\mathbb{R}}$). $\ \square$

Now it is known (p37 of here) that any Möbius transformation $T$ is conjugate to either the translation $z \mapsto z+1$ or a linear map $z \mapsto \lambda z$ for some $\lambda \in \mathbb{C} \setminus \{0\}$. Fixing $t \geq 0$ and applying this to the Möbius transformation $T$ whose restriction to $\hat{\mathbb{R}}$ coincides with $F$ (as defined in Lemma 4):

- If $h \circ T \circ h^{-1}\colon z \mapsto z+1$, then the topological circle $h(\hat{\mathbb{R}})$ is strictly invariant under $z \mapsto z+1$ and so contains $\infty$; hence $h^{-1}(\infty)$ lies in $\hat{\mathbb{R}}$ and is the unique fixed point of $F$.
- If $h \circ T \circ h^{-1}\colon z \mapsto \lambda z\,$ where $|\lambda| \neq 1$, then the topological circle $h(\hat{\mathbb{R}})$ contains both $0$ and $\infty$; hence $h^{-1}(0)$ and $h^{-1}(\infty)$ are the fixed points of $F$. If $|\lambda|>1$ then all trajectories of $F$ not starting at $h^{-1}(0)$ are attracted to $h^{-1}(\infty)$; if $|\lambda|<1$ then all trajectories of $F$ not starting at $h^{-1}(\infty)$ are attracted to $h^{-1}(0)$.
- If $h \circ T \circ h^{-1}\colon z \mapsto e^{i\mu} z\,$ where $\mu \in \mathbb{R} \setminus \{0\}$, then $h(\hat{\mathbb{R}})$ is a circle of non-zero finite radius about the origin, and so $F$ is topologically conjugate to the circle rotation by angle $\mu$.
- Finally, if $h \circ T \circ h^{-1}$ is the identity function then $F$ is the identity function.

Since $F$ is conjugate to the time-$t$ map $f$ of (1), this completes the proof of the result.