Please let me know if my proof is fine, or if you'd make some modifications.
Consider the general system of ODEs $x'=Ax+b(t)$ (1), $x(0)=x_0$ where all eigenvalues of $A$ have negative real parts and $b(t)$ is continuous for all $t$. Thus we have $\lvert e^{At}\lvert\le Me^{\gamma t}$, where $M>0$, $\gamma < 0$ for $t\geq 0$.
Now, the solution of (1) is $x(t) = e^{At}x_0 + \int\limits_0^t e^{A(t-s)}b(s)ds$, so we have:
$\lvert x(t)\lvert = \lvert e^{At}x_0 + \int\limits_0^t e^{A(t-s)}b(s)ds\lvert\leq \lvert e^{At}\lvert + \lvert \int\limits_0^t e^{A(t-s)}b(s)ds\lvert$.
Let $B = \sup\limits_{0\le s \le t}\lvert b(s) \lvert$. Moreover, since $(t-s) \geq 0$ for $0\le s \le t$, $\lvert e^{A(t-s)}\rvert \le Me^{\gamma s}$ Thus,
$\lvert e^{At}\lvert + \lvert \int\limits_0^t e^{A(t-s)}b(s)ds\lvert$ $\le \lvert e^{At}\lvert + \lvert B\int\limits_0^t e^{A(t-s)}ds\lvert\le \lvert e^{At}\lvert + \lvert B\int\limits_0^t e^{A(t-s)}ds\lvert\le Me^{\gamma t}+B\int\limits_0^t Me^{\gamma s} = Me^{\gamma t}+BM\frac{1}{\gamma}(e^{\gamma t}-1).$ Thus the solution is bounded for all $t\ge 0$.