If $m=7$ and $n=5$, then $\gcd(m,n)=1$ and $\gcd(2m-n-1,n-1)=\gcd(8,4)=4$, so it would seem the proposition is not true.
Comment added later: Someone else gave a counterexample, with $m-1$ where I had $n-1$ above. The comments below still stand. End of comment added later
Every divisor that $a$ and $b$ have in common is a divisor that $a$ and $b-a$ have in common.
Conversely, every divisor that $a$ and $b-a$ have in common is a divisor that $a$ and $b$ have in common.
Therefore the set of all common divisors of $a$ and $b$ is the same as the set of all common divisors of $a$ and $b-a$.
Therefore the greatest common divisor of $a$ and $b$ is the same as the greatest common divisor of $a$ and $b-a$, and that's why Euclid's algorithm works.
Think about applying all that to similar propositions.
Later edit: Subtracting those $1$s at the end changes things. I almost wonder if somehow a slightly different identity was intended---one that is correct. But I don't know what it would be.