Okay, I think I can show that the origin is stable below. Sorry about the messy formatting.

First, let $$ y=\begin{bmatrix}0 &1\\ 1 & 0 \end{bmatrix}\exp\left( -\begin{bmatrix} -t & 0\\ 0 &-t \end{bmatrix} - \frac{1}{\omega} \sin (\omega t) \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix}\right)x$$
and cleaning this up we have
$$y = e^{t + \frac{1}{\omega} \sin (\omega t)} \begin{bmatrix} -\sinh(\frac{1}{\omega} \sin(\omega t)) & \cosh(\frac{1}{\omega} \sin(\omega t)) \\ \cosh(\frac{1}{\omega} \sin(\omega t)) & -\sinh(\frac{1}{\omega} \sin(\omega t))\end{bmatrix} x. $$
So we are using the fundamental matrix $X(t) = e^{-t - \frac{1}{\omega} \sin (\omega t)} \begin{bmatrix} \cosh(\frac{1}{\omega} \sin(\omega t)) & -\sinh(\frac{1}{\omega} \sin(\omega t)) \\ -\sinh(\frac{1}{\omega} \sin(\omega t)) & \cosh(\frac{1}{\omega} \sin(\omega t))\end{bmatrix}$ of the solution to the original equation for $\gamma = 1$ to get a simpler equation equation for $y$, and then multiplying it by the matrix $\begin{bmatrix}0 &1\\ 1 & 0 \end{bmatrix}$ to simplify the anlysis below.

Taking the derivative of $y$ gives the messy expression:
$$ \dot y = e^{t + \frac{1}{\omega} \sin (\omega t)} \begin{bmatrix} -\sinh(\frac{1}{\omega} \sin(\omega t)) & \cosh(\frac{1}{\omega} \sin(\omega t)) \\ \cosh(\frac{1}{\omega} \sin(\omega t)) & -\sinh(\frac{1}{\omega} \sin(\omega t))\end{bmatrix} \left( \begin{bmatrix} 1+ \cos(\omega t) & 0 \\ 0 & 1+\cos(\omega t)\end{bmatrix} + \cos(\omega t)\begin{bmatrix} \sinh(\frac{1}{\omega} \sin(\omega t)) & \cosh(\frac{1}{\omega} \sin(\omega t)) \\ \cosh(\frac{1}{\omega} \sin(\omega t)) & \sinh(\frac{1}{\omega} \sin(\omega t)) \end{bmatrix} \begin{bmatrix} -\cosh(\frac{1}{\omega} \sin(\omega t)) & \sinh(\frac{1}{\omega} \sin(\omega t)) \\ \sinh(\frac{1}{\omega} \sin(\omega t)) & -\cosh(\frac{1}{\omega} \sin(\omega t)) \end{bmatrix} + \begin{bmatrix} -1 & 0 \\ 0 & -\gamma \end{bmatrix} + \cos(\omega t) \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix}\right)x,$$
where the two matrices with hyperbolic sines and cosines multiply to give $\begin{bmatrix}0 & 1 \\ 1 & 0 \end{bmatrix}$, hence
$$\dot{y} = \epsilon \begin{bmatrix} -\sinh(\frac{1}{\omega} \sin(\omega t)) & \cosh(\frac{1}{\omega} \sin(\omega t)) \\ \cosh(\frac{1}{\omega} \sin(\omega t)) & -\sinh(\frac{1}{\omega} \sin(\omega t)) \end{bmatrix} \begin{bmatrix} 0& 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} \sinh(\frac{1}{\omega} \sin(\omega t)) & \cosh(\frac{1}{\omega} \sin(\omega t)) \\ \cosh(\frac{1}{\omega} \sin(\omega t)) & \sinh(\frac{1}{\omega} \sin(\omega t))\end{bmatrix}y $$
where $\epsilon= 1-\gamma$. Multiplying this out, we finally get

$$ \dot{y} = \epsilon \begin{bmatrix} \cosh^2(\frac{1}{\omega} \sin(\omega t)) & \cosh(\frac{1}{\omega} \sin(\omega t)) \sinh(\frac{1}{\omega} \sin(\omega t)) \\ -\cosh(\frac{1}{\omega} \sin(\omega t)) \sinh(\frac{1}{\omega} \sin(\omega t)) & -\sinh^2(\frac{1}{\omega} \sin(\omega t))\end{bmatrix}y. $$

Let $$A(t) = \begin{bmatrix} \cosh^2(\frac{1}{\omega} \sin(\omega t)) & \cosh(\frac{1}{\omega} \sin(\omega t)) \sinh(\frac{1}{\omega} \sin(\omega t)) \\ -\cosh(\frac{1}{\omega} \sin(\omega t)) \sinh(\frac{1}{\omega} \sin(\omega t)) & -\sinh^2(\frac{1}{\omega} \sin(\omega t))\end{bmatrix}.$$ Note that $\epsilon = 0$ gives $y$ constant, which makes sense as it describes a perturbation of the original equation for $\gamma=1$, and further that $A(t)$ is periodic, so by Floquet's theorem we have a periodic (of period $T= \frac{2\pi}{\omega}$) matrix $P(t)$ and a constant matrix $B$ such that $y(t) = P(t)e^{tB}y(0)$ and $\text{tr}B = \epsilon$. Denote the fundamental matrix for this system by $Y(t) : = P(t)e^{tB}$.

Let $a := Y(t)a_0$ and $b := Y(-t)b_0$, for some constant vectors $a_0$, $b_0$. Since $A(-t) = A(t)^T$, we have $\dot{b}- = -\epsilon A(t)^T b $, and so the derivative of the inner product $\langle a,b\rangle$ is

$$ \frac{d}{dt} \langle a,b \rangle = \langle \epsilon A a,b\rangle + \langle a, -\epsilon A^T b\rangle = 0, $$

so $\langle a,b\rangle = \langle a_0, b_0 \rangle$. Hence, we have $P(t)e^{tB} (P(-t)e^{-tB})^T = I$, and therefore

$$e^{tB}e^{-tB^T} =P(t)^{-1} (P(t)^T)^{-1} .$$

Since both sides above must be periodic, both are equal to the identity $I$. Hence $B=B^T$ is symmetric and has real eigenvalues and eigenvectors. Since $\text{tr} B = \epsilon>0$ for $0<\gamma<1$, one eigenvalue is positive. Let us call $\lambda_+$ the largest eigenvalue of $B$, with eigenvector $v_+$. Similarly, call the other eigenvalue $\lambda_-$, with eigenvector $v_-$.

Now, note that the tangent of an integral curve (i.e. the graph of a solution) is given by $m(t):=\frac{\dot{y}_2}{\dot{y}_1} = - \tanh (\frac{1}{\omega}\sin(\omega t))$, and so $|m(t)| \leq |\tanh(\frac{1}{\omega})|<1$ for all $t \in \mathbb{R}$.

Edit to clarify this: the equations for $\dot{y}$ are

$$\dot{y}_1 = \epsilon \cosh(\frac{1}{\omega} \sin(\omega t))\left(\cosh(\frac{1}{\omega} \sin(\omega t))y_1 + \sinh(\frac{1}{\omega} \sin(\omega t))y_2 \right) $$

and

$$\dot{y}_2 = -\epsilon \sinh(\frac{1}{\omega} \sin(\omega t)) \left(\cosh(\frac{1}{\omega} \sin(\omega t))y_1 + \sinh(\frac{1}{\omega} \sin(\omega t))y_2\right).$$

Suppose $\lambda_-<0$. Then $P(t)e^{tB}v_- \to 0$ as $t \to \infty$. Now, since the slope of the integral curve is bounded, the origin must lie in the cone about a horizontal line through $v_-$ bounded by the lines of slope $\pm |\tanh(\frac{1}{\omega})|$ through $v_-$. Hence, $v_-$ lies inside a cone at the origin, which contains the $x$-axis and is bounded by lines of slope $\pm |\tanh(\frac{1}{\omega})|$ through the origin. Similarly, since $P(t)e^{tB}v_+ \to 0$ as $t\to -\infty$ and since its slope has the same bound, $v_+$ also lies in this cone at the origin. Hence, as the cone angle is less than $90^{\circ}$, the cosine of the angle between $v_+$ and $v_-$ is nonzero. So consider the solution $w(t):= P(t)e^{tB}(v_++ v_-)$. We now have $$\| v_+\|^2 +2 \langle v_+,v_-\rangle + \|v_-\|^2=\langle w(nT),w(-nT) \rangle = \langle e^{\lambda_+ nT}v_+ + e^{\lambda_- n T}v_-, e^{-\lambda_+ nT}v_+ + e^{-\lambda_- n T}v_- \rangle = \|v_+\|^2+2\cosh((\lambda_+ - \lambda_-)nT)\langle v_+,v_- \rangle+\|v_-\|^2$$
and so, since $\lambda_+ \neq \lambda_-$, we must have $\langle v_+,v_- \rangle=0$, a contradiction. Hence, $\lambda_- \geq 0$.

We are now done, as if both eigenvalues are positive and satisfy $\lambda_+ + \lambda_- = \epsilon$, both are smaller than $\epsilon$. So a solution $x$ to the perturbed system for $0<\gamma<1$ satisfies $\|x\| \leq e^{-t}e^{(1-\gamma)t}\|Q(t)x_0\|$, where $Q(t)$ is some periodic matrix, and hence $x(t) \to 0$ as $t \to \infty$.

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