As noted above (and standard anyway) $M(t)=E(e^{tx})=\int_1^{oo} e^{tx}e^{-(x-1)} dx$. Note that if t=1 the integrand becomes simply e^1, and the integral does not exist. So M(1) is undefined. Also note that if $t>1$ the integrand is $e^{(t-1)x+1}>e^{(t-1)x}$ which since $t>1$ goes to infinity as $x$ does. So M(t) does not exist for t>1.
We claim now that M(t) exists for t<1. The antiderivative of the integrand is $\frac {e^{-x(1-t)}e^1}{t-1}$, which goes to $0$ as $x$ approaches infinity, and evaluates to $\frac {e^t}{t-1}$ at $x=1$. We thus have the formula
$M(t)=\frac {e^t}{1-t}$
for the moment generating function, on the domain $\{t:t<1\}$. Note that this domain includes an open neighborhood of 0, so that moment generating calculations of moments can be done as usual.
To get the mean, second moment, and variance we only need a few terms of the series for $M(t)$. We can get these by multiplying the series for $e^t$ by the series for $1/(1-t)$. That is,
$(1+t+t^2/2+t^3/6+...)(1+t+t^2+t^3+...)$
which gives first few terms $M(t)=1+2t+(5/2)t^2+(8/3)t^3+...$. Then the usual procedure of taking derivatives and putting in zero gives $E(X)=2$ and $E(X^2)=5$. Finally, using the standard formula for variance we get $V(X)=E(X^2)-[E(X)]^2=5-2^2=1$.
To complement the usual approach, note that $X=UY$ where $U$ and $Y$ are independent, $Y$ is a standard exponential random variable and $\mathbb P(U=\frac12)=\mathbb P(U=-1)=\frac12$.
This yields readily the moment generating functions. For example, conditioning on $U$,
$$
M_X(t)=\mathbb E(\mathrm e^{tX})=\mathbb E(\mathbb E(\mathrm e^{tX}\mid U))=\mathbb E(M_Y(tU)).
$$
Since $M_Y(t)=1/(1-t)$ for every $t\lt1$, this yields, for every $-1\lt t\lt2$,
$$
M_X(t)=\mathbb E\left(\frac1{1-tU}\right)=\frac12\frac1{1-\frac12t}+\frac12\frac1{1+t}.
$$
Likewise, for every $t\lt1$,
$$
M_{|X|}(t)=\mathbb E(\mathrm e^{t|X|})=\mathbb E\left(\frac1{1-t|U|}\right)=\frac12\frac1{1-\frac12t}+\frac12\frac1{1-t}.
$$
Best Answer
We begin, as you indicated, with the integral $\int_{-\infty}^\infty e^{tx}f(x)dx$ In order to find this integral, we may proceed as follows:
$$ \int_{-\infty}^\infty e^{tx}f(x)dx = \int_0^\infty x e^{tx} e^{-x}dx =\int_0^\infty x e^{(t-1)x}dx $$ From there, use integration by parts. That is, we have $\int u\,dv = uv - \int v\, du$. For this problem, we choose $u = x$ and $dv = e^{(t-1)x}dx$. Applying the rule gives us $$ \begin{align} \int_0^\infty xe^{(t-1)x}dx &= \left[\frac{1}{t-1}x e^{(t-1)x}\right]_0^\infty - \int_0^\infty \frac{1}{t-1} e^{(t-1)x}dx\\ &= \frac{1}{t-1} \left[x e^{(t-1)x}\right]_0^\infty - \frac{1}{(t-1)^2}\left[e^{(t-1)x}\right]_0^\infty\\ &= \frac{1}{t-1} \cdot 0 - \frac{1}{(t-1)^2}\cdot (-1)\\ &= \frac{1}{(t-1)^2} \end{align} $$