The answer is yes. Indeed, the first and second displayed inequalities in the OP yield the following for $t\ge1$: \begin{align*} x(t)&=x(1)+\int_1^t du\, x'(u) \\ &\le x(1)+\int_1^t du \int_u^\infty ds\, x(s)\frac{k(s)}{s^2} \\ &= x(1)+\int_1^\infty ds\, x(s)\frac{k(s)}{s^2}\int_1^{\min(s,t)} du \\ &\le x(1)+\int_1^\infty ds\, x(s)\frac{k(s)}s<\infty, \end{align*} as desired.

Without requiring that the integral \begin{equation} \int_1^\infty x(s)\frac{k(s)}{s^2}\,ds \tag{1} \end{equation} be finite, the answer is no. Indeed, then one can take e.g. $k(s)=1/s$ and $x(s)=e^s$.

I don't know whether the condition that the integral in (1) be finite changes the answer.