For any $a > 0$ take $\vp(t)=(t-a)^+$. What inequality can be deduced? Find an optimal value for $a$ and show that
$$
\Big(\Big(\int f^*\,d\mu\Big)^{1/2}-\sqrt{\mu(S)/e}\Big)^2
\leq\int f\log^+f\,d\mu+\mu(S)/e~.
$$
$\vp^\prime=I_{(a,\infty)}$: $\Phi(t)=\log^+(t/a)$
$$
\int f^*\,d\mu
\leq\int(f^*-a)^+\,d\mu+a\mu(S)
\leq\int\log^+(f^*/a)f\,d\mu+a\mu(S)
\leq\int f\log^+f+(ae)^{-1}f^*\,d\mu+a\mu(S)
$$
Now $\inf_{a > 0}A/a+aB=2\sqrt{AB}$ and thus for $I\colon=\int f^*\,d\mu$ and
$A=I/e$, $B=\mu(S)$:
$$
(\sqrt I-\sqrt{\mu(S)/e})^2
=I-2\sqrt{I\mu(S)/e}+\mu(S)/e
\leq\int f\log^+f\,d\mu+\mu(S)/e
$$
It follows that
$$
I\leq\Big(\sqrt{\int f\log^+f\,d\mu+\mu(S)/e}+\sqrt{\mu(S)/e}\Big)^2
\leq2\int f\log^+f\,d\mu+4\mu(S)/e~.
$$