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.