$\newcommand\la\lambda\newcommand{\R}{\mathbb R}\newcommand{\de}{\delta}$The answer to this question is yes. 

Indeed, 
let $h:=f^*/g$. Then $h$ is a nonincreasing function and the inequality in question can be rewritten as 
\begin{equation*}
	\Big(\int_0^\infty h(s)^{p'}ds\Big)^{p/p'}
\le c\int_0^\infty h(s)^p s^{p/p'-1}\,ds. \tag{1}\label{1}
\end{equation*}
Note that $p'>p>0$. 

Letting 
\begin{equation*}
	a:=p'/p>1\quad\text{and}\quad u:=h^p,
\end{equation*}
we see that it is enough to show that for all real $N>1$ 
\begin{equation*}
	L^{1/a}\overset{\text{(?)}}\le2R, \tag{2}\label{2}
\end{equation*}
where 
\begin{equation*}
	L:=\int_0^N u(s)^a\,ds,\quad 
	R:=\int_0^N u(s)s^{1/a-1}\,ds,
\end{equation*}
and $u\colon[0,N]\to\R$ is a nonnegative nonincreasing function. 

If $R=0$, then $L=0$ and hence inequality \eqref{2} is trivial. So, in what follows assume that $R>0$. 

Then, by "vertical" rescaling, without loss of generality (wlog) $u(0)=1$. 

The set, say $U_N$, of all nonnegative nonincreasing functions $u$ on the interval $[0,N]$ with $u(0)=1$ is sequentially compact in the topology of almost everywhere convergence. So, there is a maximizer of $L$ over all $u\in U_N$ with a fixed positive value of $R$. 

In what follows, wlog let $u$ be such a maximizer. Let 
\begin{equation*}
	\de:=\max\{t\in[0,N]\colon u=1\text{ on }[0,t)\}, 
\end{equation*}
\begin{equation*}
	T:=\max\{t\in[0,N]\colon u>0\text{ on }[0,t)\}. 
\end{equation*}
Note that $0\le\de\le T\le N$, $u=1$ on $[0,\de)$, $0<u<1$ on $(\de,T)$, and $u=0$ on $(T,N]$. 

If $\de=T$, then $L=\de$ and $R=a\de^{1/a}$, so that \eqref{2} holds trivially. So, wlog $\de<T$. 

Using now Lagrange multipliers, we see that for some real $\la$ and almost all $s\in(\de,T)$ we have $u(s)^{a-1}=\la s^{1/a-1}$; it also follows that $\la>0$. So, 
\begin{equation*}
	u(s)=b s^{-1/a}  \tag{5}\label{5}
\end{equation*}
for some real $b>0$. Wlog, \eqref{5} holds for all $s\in(\de,T)$. So, $b\de^{-1/a}=u(\de+)\le u(0)=1$ and hence 
\begin{equation*}
	b\de^{-1/a}\le1.  \tag{6}\label{6}
\end{equation*}

If now $\de=0$, then $R=\infty$, so that \eqref{2} holds trivially again. So, wlog $\de>0$.  

Now \eqref{2} becomes 
\begin{equation*}
	(\de+b^a l)^{1/a}\le2(a\de^{1/a}+bl),
\end{equation*}
where $l:=\ln(T/\de)>0$. Therefore and because $a>1$, it is enough to show that 
\begin{equation*}
	\de^{1/a}+b l^{1/a}\le2(a\de^{1/a}+bl). \tag{7}\label{7}
\end{equation*}
If $l\ge1$, then $l^{1/a}\le l$, so that \eqref{7} follows. If $l\le1$, then 
$b l^{1/a}\le b\le\de^{1/a}$ by \eqref{6}, 
so that \eqref{7} again follows. $\quad\Box$ 

---

Working slightly harder at the end of the above proof, we can see that \eqref{1} holds with $c=1/a<1$.