To show that $(\alpha,\omega)$ is a fibration, we must define a path-lifting function $L$ as follows. The arguments are a length $d\geq 0$, a Moore path $u:[0,d]\to X$, a path $v:[0,1]\to X$ starting at $u(0)$, and a path $w:[0,1]\to X$ starting at $u(d)$. The output $L(d,u,v,w)$ must be a path in $MX$ such that $L(d,u,v,w)(t)$ is a Moore path from $v(t)$ to $w(t)$. This is easy to arrange: we construct $L(d,u,v,w)(t)$ by gluing the reverse of $v|_{[0,t]}$ with $u$ and $w|_{[0,t]}$.
Also, there is an evident inclusion $f:X^I\to MX$, and a map $g:MX\to X^I$ given by $g(d,u)(t)=u(dt)$. Both of these are compatible with $\alpha$ and $\omega$. We have $gf=1$ on the nose. I believe that $fg$ is homotopic to the identity, but I don't have time to check that right now.