The boundedness statement is true
The general argument is similar to what I gave in my previous answer, which was essentially off by a log due to certain inefficiencies in the estimates. Here I rewrite the argument to get rid of the log loss.
To start, as explained in the previous answer, we immediately have that $x(t)$ is sublinear. Since the inequality concerning $x(t)$ is linear, we can assume, without loss of generality, that $\sup x(t)/t \leq 1$ by a simple rescaling of $x$. This means we can find a sequence of times $1 = t_0, t_1, t_2, \ldots$ defined by
$$ t_i = \inf \{ t\in [1,\infty) : \forall s \geq t, x(s) \leq 2^{-i} s \}. $$
(Note that implicitly since $x$ is differentiable it is also continuous, so $x(t_i) = 2^{-i} t_i$.)
Our goal is to estimate $t_i$. Specifically, we want to show that $2^{-i}t_i$ is bounded.
We will also denote by
$$ K(t) = \int_t^\infty \frac{k(s)}{s} ~ds, \quad K_i = \int_{t_i}^{t_{i+1}} \frac{k(s)}{s} ~ds.$$
We note that $K_i \searrow 0$ and the numbers are in fact summable by assumption.
Gronwall
Integrating by parts the differential inequality for $x'$ we get (as I argued in the previous answer) for $1 \leq a < b$
$$ x(b) - x(a) \leq (b-a) \int_b^\infty \frac{x(s)}{s} \frac{k(s)}{s} ~ds + \int_a^b x(s) \frac{s - a}{s} \frac{k(s)}{s} ~ds $$
Estimating $(s-a)/s \leq 1$ we have that, by Gronwall's inequality
$$ x(b) \leq \left[ x(a) + (b-a) \int_b^\infty \frac{x(s)}{s} \frac{k(s)}{s} ~ds\right] \cdot e^{K(a) - K(b)} $$
This implies, setting $b = t_{i+1}$ and $a = t_i$, that
$$ 2^{-1-i} t_{i+1} \leq \left[ 2^{-i} t_i + (t_{i+1} - t_i) \sum_{j = i+1}^\infty 2^{-j} K_j \right] e^{K_{i}} $$
(here we rewrote $\int_{t_{i+1}}^\infty x(s) k(s) s^{-2} ~ds = \sum_{j = i+1}^\infty \int_{t_j}^{t_{j+1}} x(s)s^{-1} \cdot k(s) s^{-1} ~ds$, and used the decaying bound on $x(s)s^{-1}$ above $t_j$, and the fact that all functions involved are positive.)
Simplify (by the summability of $K_j$ we can assume from here on the indices $i$ are always larger than some sufficiently large $i_0$ such that the two terms in the brackets below are guaranteed to be positive)
$$ \left[ e^{-K_{i}} - \sum_{j = 0}^\infty 2^{-j} K_{i+1+j} \right] t_{i+1} \leq \left[2 - \sum_{j = 0}^\infty 2^{-j} K_{i+1+j} \right] t_i $$
So for all sufficiently large $i$ we have the bound (using the convexity of the exponential function)
$$ t_{i+1} \leq 2 t_i \cdot \frac{1}{e^{-K_{i}} - \sum_{j = 0}^\infty 2^{-j} K_{i+1+j}} \leq \frac{2 t_i}{1 - K_i - \sum_{j = 0}^\infty 2^{-j} K_{i + 1 + j}} $$
The Estimates on $t$
To show our desired conclusion it suffices to show that the infinite product
$$ \prod_{i = i_0}^\infty (1 - K_i - \sum_{j = 0}^\infty 2^{-j} K_{i+1+j}) $$
is bounded below away from zero. Now,
$$ \sum \ln (1 - K_i - \sum_{j = 0}^\infty 2^{-j} K_{i+1+j}) \geq - \sum_{i = i_0}^\infty \left( K_i + \sum_{j = 0}^\infty 2^{-j} K_{i + 1 + j} \right) $$
by convexity, the first term in the sum is obviously bounded by the summability of $K_i$. For the second term, using the monotonicity of $K_j$ (which follows from the monotonicity of $k(s)$) we have
$$ \sum_{i = i_0}^\infty \sum_{j = 0}^\infty 2^{-j} K_{i + 1 + j} \leq \sum_{i = i_0}^\infty \sum_{j = 0}^\infty 2^{-j} K_{i+1} $$
and is also obviously bounded.
This concludes the proof. In fact, as a consequence of this argument, we can have a quantitative bound on $x$.
Suppose we know that $\int_1^\infty \frac{k(s)}s ~ds =:\kappa < 1$ (we can get a similar statement if we shift from $1$ to $t_{i_0}$ with an appropriate factor inserted). Then the above argument in fact implies that
$$ x(t) \leq x(1) \cdot e^{3\kappa}.$$