Counter-example. Let $\left(z(t)\right)_{t\geq 0}$ be the solution to
$$\dot{z}(t)=A z(t)-\varepsilon \mathbf{e}_2,$$
with $\mathbf{e}_2:=\left[0 \,\,\,\,1\right]^{\top}$ and initial condition $z(0)$. Then, $\left(z(t)\right)_{t\geq 0}$ obeys the inequality $\dot{z}(t)\leq Az(t)$ for all $t$.
Let $\left(x(t)\right)_{t\geq 0}$ be the solution to $\dot{x}(t)=Ax(t)$ with initial condition $x(0)=z(0)$. Then, $x(t)=e^{At}z(0)$ for all $t\geq 0$.
Now, remark that,
$$ \dot{z}_1(0)=\left[Az(0)\right]_1=\left[Ax(0)\right]_1=\dot{x}_1(0)$$
and
$$ \ddot{z}_1(0)=-\beta \dot{z}_2(0)>-\beta \dot{x}_2(0)=\ddot{x}_1(0),$$
since $\dot{z}_2(0)=\left[Az(0)\right]_2-\varepsilon < \left[Az(0)\right]_2 =\left[Ax(0)\right]_2 = \dot{x}_2(0).$
In other words, $w(t)\overset{\Delta}= z_1(t)-x_1(t)$ is analytic with $w(0)=0$, $\dot{w}(0)=0$ and $\ddot{w}(0)>0$. Therefore, $w(t)>0$ for all $t\in\left(0,\delta\right)$, for some $\delta>0$ small enough. That is,
$$z_1(t)>\left[e^{At}z(0)\right]_1$$
for all $t\in\left(0,\delta\right)$ which breaks Grönwall's.
For the record. If the two off-diagonal entries of $A$ are negative (with arbitrary diagonal entries), then we can show that Grönwall's inequality holds w.r.t. the partial order $\leq_a$ defined as $x\leq_a y \Leftrightarrow (x_1 \geq y_1 \wedge x_2 \leq y_2)$
or to the partial order $\leq_b$ defined as $x\leq_b y \Leftrightarrow (x_1 \leq y_1 \wedge x_2 \geq y_2)$. In other words, for this regime, if $\dot{z}(t)\leq_i Az(t)$ then, $z(t)\leq_i e^{At} z(0)$ for $i\in \left\{a,b\right\}$.
This is done simply via comparing $\left(z(t)\right)$ and $\left(x(t)\right)$ (where $x(t)$ is the solution to the linear system with $x(0)=z(0)$) at the hitting moment $T$ for the possible cases: i) $z_1(T)=x_1(T)$ and $z_2(T)<x_2(T)$; ii) $z_1(T)>x_1(T)$ and $z_2(T)=x_2(T)$; or iii) $z_1(T)=x_1(T)$ and $z_2(T)=x_2(T)$. In the latter case you need to resort to the higher order derivatives as $\dot{z}_1(T)=\dot{x}_1(T)$ and $\dot{z}_2(T)=\dot{x}_{2}(T)$.
