There is nothing incorrect, but there are a few things that could be changed. We only need $p>2$.
From $2^p \equiv 2 \pmod {p}$ one should conclude $M_p=2^p -1\equiv 1 \pmod{p}$ immediately, without the detour through $a\cdot p +1$.
A similar needless detour is made when from $M_p\equiv 7\pmod{24}$ it is argued that $M_p \equiv 1 \pmod{p}$. Instead, if we are going to go that route, we could note that since $6$ is a factor of $24$, we have $M_p\equiv 7\equiv 1\pmod{6}$. But the assertion $M_p \equiv 7 \pmod{24}$ may not be familiar to people, and it is easy to prove. We won't bother, and will only prove what you need.
To prove your result, all you need is to show that $M_p \equiv 1 \pmod{6}$. It is clear that $M_p$ is odd, that is, $M_p\equiv 1 \pmod{2}$. So all we need is to show that $M_p \equiv 1 \pmod 3$. We have $2\equiv -1\pmod 3$. So if $n$ is odd, then $2^n\equiv (-1)^n \pmod{3}$, and therefore $M_p=2^p-1\equiv (-1)-1\equiv 1\pmod {3}$.
It is a standard fact about congruences that if $x\equiv a \pmod{m}$ and $x\equiv a \pmod{n}$, then $x\equiv a \pmod{\text{lcm}(m,n)}$. You chose to prove it in the special case $m=6$, $n=p$. That's fine, perhaps not needed.
Comment: The congruence machinery is very powerful, and it is useful to learn to exploit it efficiently.