Given $x^9 = e$ and $x^{11} = e$ prove $x = e$.

Your reasoning is correct but here is a more direct argument.

Since $x^9 = e$ and $x^{11} = e$, the order of $x$ divides both $9$ and $11$. Therefore, the order of $x$ is $1$ so $x = e$.

$$e=(x^{11})^5=x^{55}=x^{54}\cdot x=(x^9)^{6}\cdot x=e\cdot x=x$$

As you can see, this thus follows because there is integer solutions to $11x-9y=1$, which is true because $11$ and $9$ are relatively prime.

Your approach is doing much the same, using a slow form of the Euclidean algorithm to show that $11$ and $9$ are relatively prime:

$$11=9\cdot 1 + 2\\ 9=2\cdot 1 + 7\\ 7=2\cdot 1 + 5\\ 5=2\cdot 1 + 3\\ 3=2\cdot 1 + 1$$

You could have skipped a lot of steps by doing the equivalent of $9=2\cdot 4 + 1$, as other answers have suggested.

Yes, it is that simple. It can be done even shorter, because after showing $x^2=e$, you can go straight to $$e=x^9=x(x^2)^4=xe^4=x$$and you're done.