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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.