For every $s\gt0$, let $\nu(s)=\mu(s)/s+P(X\gt s)$, then $\nu(s)=E(u_s(X))$ where, for every $x\geqslant0$, $u_s(x)=\min\{1,x/s\}$. Assume that $P(X=0)\ne1$. Then, for every $x\geqslant0$, $u_s(x)\leqslant1$, $u_s(x)\leqslant u_t(x)$ for every $s\geqslant t$, $u_s(x)\to0$ when $s\to+\infty$, and, for every $x\gt0$, $u_s(x)\to P(X\gt0)$ when $s\to0$.
Thus, the function $\nu$ is nonincreasing and, by dominated convergence, $\nu(s)\to P(X\gt0)$ when $s\to0$ and $\nu(s)\to0$ when $s\to+\infty$.
Furthermore, for every $s\leqslant t$, $\|u_s-u_t\|_\infty=u_s(s)-u_t(s)=(t-s)/t$ hence the function $\nu$ is continuous.
If the distribution of $X$ is continuous, the function $s\mapsto\mu(s)/s=\nu(s)-P(X\gt s)$ is continuous, with limits $0$ when $s\to0$ and when $s\to+\infty$. Furthermore, there exists some $s^*\gt0$ such that $P(0\lt X\lt s^*)\gt0$, hence $\mu(s^*)\gt0$. Let $M=\mu(s^*)/s^*$, then, for every $x$ in $(0,M)$ there exists some $\beta(x)$ in $(s^*,+\infty)$ such that $\mu(\beta(x))/\beta(x)=x$.
For every $n$ large enough, $1/n\lt M$ hence $b_n=\beta(1/n)$ solves the question in this case.
If the distribution of $X$ has some atoms, the continuity argument used above fails (the function $\nu$ is still continuous but not the function $\mu$) and actually, one doubts that the result holds in full generality.
