I want to find $M(t)$ of
$$f(x)= \begin{cases} e^{(-x-1)} & \text{for } x > -1 \\ 0 & \text{otherwise} \end{cases}$$
$e^{(-x-1)}$
I tried to do $$\int_{-1}^{∞ {}} e^{tx} e^{-x-2}\,dx.$$
[integral of e^(tx)*e^(-x-1) from -1 to infinity]
but don't believe that is correct.
What you are doing is correct, except there was a typo where the '2' should be a '1'.
$$ \begin{aligned} M(t)&=E(e^{tx})=\int_{-1}^{\infty} e^{tx} e^{-x-1} dx \\ &=e^{-1}\int_{-1}^{\infty} e^{(t-1)x} dx \\ &=e^{-1}\frac{1}{t-1}\left. e^{(t-1)x} \right|_{-1}^\infty \\ &=e^{-1}\frac{e^{1-t}}{1-t} \\ &=\frac{e^{-t}}{1-t} \\ \end{aligned} $$ so long as $t<1$.
We can do a quick check: The distribution given is a standard exponential distribution with parameter $1$ but shifted $1$ unit in the negative direction and thus we can argue that $E(X+1)=1 \Rightarrow E(X)=0$. The mean should be given by $E(X)=M'(0)=0$ which checks out. Now you can take a few more derivatives of $M(t)$ and plug in $t=0$ and calculate the moments of $X$ directly by integration: $E(X^n)$ and compare.