This is not possible (updated question). If $0\lt p\lt \infty$, $f$ is in $L^p$, and $Mf$ is in $L^p$, then $f$ is zero a.e.. In other words, if $0\lt\int_\mathbb{R}|f|^p\lt\infty$, then $\int_\mathbb{R}|Mf|^p=\infty$ (and in fact, |Mf| has a positive lower bound outside of some bounded interval).
Suppose that $0\lt\|f\|_p\lt\infty$. There is some bounded interval $[a,b]$ with $\int_a^b|f|^p\gt0$. WLOG (rescale and shift) assume $[a,b]=[-1,0]$. Thus the essential supremum of $|f|$ on $[-1,0]$ is positive, so there is a positive number $c$ and a set $E\subseteq[-1,0]$ of positive measure $m$ such that $|f(x)|\gt c$ for all $x\in E$. There is a $K\gt 0$ such that the measure of the set $\{x>K:|f(x)|\geq \frac{c}{2}\}$ is less than $\frac{m}{2}$. For $x\gt K/2$, as $t$ ranges over the interval $[x,x+1]$, $x-t$ ranges over $[-1,0]$ while $x+t$ is contained in $(K,\infty)$. Thus the set of such $t$ with $|f(x-t)|>c$ and $|f(x+t)|<\frac{c}{2}$ has measure greater than $\frac{m}{2}\gt 0$ (and in particular is nonempty, but this would also work if you changed the sup to an essential sup). Therefore $|Mf(x)|\gt \frac{c}{4}$ for all $x\gt K/2$.