An infinite subsequence of random numbers in Kolmogorov sense
If $x$ is Kolmogorov random then $C_U(ax+b) \geq \log x - O(\log a + \log b + 1)$ (why?), and so
$$
\frac{C_U(ax+b)}{\log (ax+b)} \geq \frac{\log x - O(\log a + \log b + 1)}{\log x + O(\log a + \log b + 1)} = 1 - O\left(\frac{\log a + \log b + 1}{\log x}\right).
$$
Since there are infinitely many Kolmogorov random numbers $x$, your result follows.