I don't think you need Poincare Bendixson here. This is an integrable system, it has a first integral. Here is what you do.
Observe that all curves $\{(x_0 - t, 0) : t \in \mathbb{R}\}$ starting from a point $(x_0, 0)$ on the $x-$axis trace the $x-$axis, so your phase space, after removal of the orbit $\{(x,y) \, \in \, \mathbb{R}^2 \, : \,\, y=0\}$, is divided into an upper half-plane and a lower half plane.
Next, for $y>0$ (upper half-plane) change the variables as follows: $x = x \, , \,\,\, y = e^z$. Then
$\dot{y} = e^z \, \dot{z} = -x \, y = - x \, e^z$ which after canceling out the term $e^z$ on both sides of the latter equality, leads to the system
$$
\begin{align}
\dot{x} &= e^z - 1\\
\dot{z} &= -x
\end{align}
$$ Then you cast this system in the form of a second order (Newton's type) equation
$$\ddot{z} = - \dot{x} = 1 - e^{z}$$ i.e.
$$\ddot{z} = 1-e^z$$
This has a conservation of energy integral of motion
$$\frac{\dot{z}^2}{2} + e^z - z = E_0$$ for a constant $E_0$. Here $\frac{\dot{z}^2}{2}$ is the kinetic energy term and $U(z) = e^z-z$ is the potential energy term. Then when you draw the graph of the $U(z) = e^z-z$ you see that it is a convex function with exactly one minimum, equal to $1$ when $z=0$, and growing to $+\,\infty$ when $z \to \pm \, \infty$. The latter means that for each energy level $E_0 > 1$ we have a periodic orbit circling round the equilibrium $x=0, z = 0$ which is the equilibrium $x=0, y=1$ in the original coordinates. Hence, the whole upper half-plane is filled with periodic solutions.
For the lower half-plane $y<0$, change the variables as follows: $x=x\, , \,\, \, y= - e^{z}$. Then proceed analogously. You end up with the second order equation
$$\ddot{z} = e^{z} + 1$$ with a conserved total energy
$$\frac{\dot{z}^2}{2} - \big(z + e^{z}\big) = E_0$$ If you draw the graph of the potential $U(z) = - \big(z + e^{z}\big)$ you observe that it is a strictly decreasing function so the energy level for each $E_0$ are semi-open trajectories (open on one side, going to infinity). In particular they are never periodic solution.
You're almost done, you just need to integrate :
$$x(T)-x(0) = \int_0^T \frac{dE(x(s))}{ds} ds = \int_0^T |\nabla E (x(s))|^2 ds,$$
so if $x(T) = x(0)$, then the integral is zero; since the integrand is nonnegative, is must be also be zero.
Best Answer
Let $\psi(t) : \mathbb{R} \to L(\mathbb{R}^n, \mathbb{R}^n)$ denote the ordered exponential of $A(t)$, i.e., $\psi(t)$ is the unique solution of
$$ \psi'(t) = A(t)\psi(t), \qquad \psi(0) = 1. $$
Then
\begin{align*} (\det\psi(t))' &= \lim_{\Delta t \to 0} \frac{\det\psi(t+\Delta t) - \det\psi(t)}{\Delta t} \\ &= \lim_{\Delta t \to 0} \frac{\det\left(1 + A(t)\Delta t + \mathcal{O}(\Delta t^2)\right) - 1}{\Delta t} \det \psi(t) \\ &= \left( \operatorname{Tr}A(t) \right) \det\psi(t). \end{align*}
So it follows that
$$ \det\psi(t) = \exp\left\{ \int_{0}^{t} \operatorname{Tr}A(s) \, \mathrm{d}s \right\}. $$
Since $y(t) = \psi(t)y_0$ solves $y'(t) = A(t)y(t)$ for any $y_0 \in \mathbb{R}^n$, $y(t)$ is $T$-periodic for any initial condition $y_0$, and so, $\psi(t)$ itself is $T$-periodic. Then from $\psi(T) = \psi(0)$, the desired conclusion follows.