A matrix monotonicity question

The answer is no, in general. Here is a counterexample:

Let \begin{align*} X = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \quad \text{and} \quad A = \begin{pmatrix} -1 & -1 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} - I, \end{align*} where $I \in \mathbb{R}^{2 \times 2}$ denotes the identity matrix. Note that the eigenvalues of $A$ are $-1\pm i$, so $A$ is stable. For every $t \in [0,\infty)$ and every $k \in \mathbb{N}_0$ we have \begin{align*} e^{tkA} = e^{-kt} \begin{pmatrix} \cos(kt) & -\sin(kt) \\ \sin(kt) & \cos(kt) \end{pmatrix}. \end{align*} Choose $y = (0,1) \in \mathbb{R}^2$. Then we have \begin{align*} \langle y, e^{tkA} X e^{tkA^T}y\rangle = \langle y, e^{tkA}X(e^{tkA})^T y\rangle = e^{-2kt}\sin^2(kt), \end{align*} for all $t \in [0,\infty)$ and $k \in \mathbb{N}_0$, so \begin{align*} \langle y, \sum_{k\ge 0} e^{tkA} X e^{tkA^T} y\rangle = \sum_{k \ge 0} e^{-2kt} \sin^2(kt) =: \varphi(t). \end{align*} for all $t \in [0,\infty)$. Now, set $t_1 := \pi$ and $t_2 := \frac{3}{2}\pi$. Then $\varphi(t_1) - \varphi(t_2) = -\varphi(t_2) < 0$, so the matrix \begin{align*} \sum_{k\ge 0} e^{t_1kA} X e^{t_1kA^T} - \sum_{k\ge 0} e^{t_2kA} X e^{t_2kA^T} \end{align*} is not positive semi-definite.