Yes we can!
There's possibly a much simpler proof. The argument below starts with the observation that if $\int_1^\infty x(s) k(s) s^{-2} ds$ converges, then $x(t)$ must be automatically sublinear. This can then be upgraded to show that $x(t)$ grows slower than any power of $t$ using Gronwall. And finally a finer analysis of the error term gives you boundedness. The key step is the Gronwall analysis.
Sublinear growth
First one observes that if the integral is finite, the growth rate is sublinear. This follows from the fact that if the integral $\int_1^\infty x(s) k(s) s^{-2} ds$ converges, then $x'(t) \to 0$ as $t\to\infty$, and hence $x(t) = o(t)$ by, e.g., L'Hopital.
Upgrade the growth control
Integrating the bound for $x'(t)$ gives, for arbitrary $1 \leq t_1 < t_2$
$$ x(t_2) - x(t_1) \leq \int_{t_1}^{t_2} \int_t^\infty x(s) k(s) s^{-2} ds~dt $$
Integrating in parts in $t$, using that $dt = d(t - t_1)$
$$ x(t_2) - x(t_1) \leq (t_2 - t_1) \int_{t_2}^\infty x(s) k(s) s^{-2} ds + \int_{t_1}^{t_2} (t - t_1) x(t) k(t) t^{-2} dt $$
Now, for any $\delta > 0$, there exists $\tau_{\delta}$ such that for every $s \geq \tau_{\delta}$, it holds that
$$ x(s) \leq \delta s, \qquad \text{and} \qquad \int_{s}^\infty k(s)s^{-1} ds < \delta $$
since we know that $x$ grows sublinearly and the relevant integral converges. This implies that, for $t_2 > \tau_{\delta}$,
$$ x(t_2) \leq x(\tau_{\delta}) + (t_2 - \tau_{\delta}) \delta^2 + \int_{\tau_{\delta}}^{t_2} x(s) \cdot k(s) s^{-1} ds $$
By Gronwall's inequality this means that
$$ x(t_2) \leq [ \delta \tau_{\delta} + (t_2 - \tau_{\delta}) \delta^2 ] e^{\delta} $$
This final inequality can be used to estimate $\tau_{\delta/2}$. Consider the inequality
$$ [\delta \tau_{\delta} + (s - \tau_{\delta}) \delta^2] e^{\delta} \leq \frac{\delta}{2} s $$
Solving this inequality we see that this is satisfied whenever
$$ \frac{e^{\delta} \tau_{\delta}}{\frac12 - \delta e^{\delta}} \leq s $$
For convenience write
$$ \frac{e^\delta}{\frac12 - \delta e^\delta} = 2^{1+\sigma(\delta)} $$
and note that $\sigma$ is continuous and $\sigma(0) = 0$.
This implies that $\tau_{\delta/2} \leq 2^{1 + \sigma(\delta)} \tau_{\delta}$.
Now, iterating this argument, starting from some sufficiently small $\delta$ that we fix, we see can construct a sequence of times $\tau_{2^{-k} \delta}$ such that for $T \geq \tau_{2^{-k} \delta}$ it holds
$$ x(T) \leq 2^{-k} \delta T $$
We also have by iteration the bounds
$$ \tau_{2^{-k}\delta} \leq 2^k 2^{\sum_{j = 0}^{k-1} \sigma(2^{-j} \delta)} \tau_{\delta} $$
Using that $\sigma$ is continuous and $\sigma(0) = 0$, this immediately implies that $x(t)$ must grow slower than any power of $t$.
Boundedness
To get boundedness, we need to control the sum
$$ \sum_{j = 0}^{k-1} \sigma(2^{-j} \delta) $$
If this sum is uniformly bounded in $k$, then we have the uniform bound $\tau_{2^{-k}\delta} \leq C 2^{k} \tau_{\delta}$, and this will imply the uniform bound for $x$ of the form $x(T) \leq C \delta \tau_{\delta}$.
Returning to the definition of $\sigma$ we see that
$$ \sigma(\delta) = \frac{1}{\ln 2} ( \delta - \ln(1 - 2\delta e^{\delta}) ) $$
Thus for all sufficiently small $\delta$ we have
$$ \sigma(\delta) \leq \frac{1}{\ln 2} ( \delta + 2 \delta e^{\delta}) \leq \frac{4}{\ln 2} \delta $$
This implies the summability of $\sigma(2^{-j} \delta)$, and the desired conclusion.