Suppose the random variable $X$ has a exponential distriubtion with parameter $1$.
Let $X^* = X - 2$ when $X\geq 2$ and $0$ otherwise. I am trying to prove using MGFs that $X^*$ and $X$ have the same distribution exponential distribution with parameter 1. I have done this using conditional probablilities and showed the CDFs are equal, but not sure how to do it using MGfs
Probably, it is assumed that $X^*$ is some random variable that is defined on the event $A=\{X\geq 2\}$ only and whose distribution is conditional on this event: $$\mathbb P'(X^*\in B)=\mathbb P(X-2\in B|X\geq 2),\ $$ where $P'$ is a new probability measure on $P$-measurable subsets of $A$, definded as $P'(C)=P(C|A)$, $C\subseteq A$?
In this case the MGF's of $X$ and $X^*$ are really coincide.
For $t<1$ $$M_X(t)=\mathbb Ee^{tX} = \int_0^\infty e^{tx}e^{-x}\,dx = \int_0^\infty e^{(t-1)x}\,dx =\dfrac{1}{t-1}.$$ Let $\mathbf 1_{\{X\geq 2\}}(\omega)=1$ if $X(\omega)\geq 2$, and $0$ otherwise. For $t<1$ $$M_{X^*}(t)=\mathbb Ee^{tX^*} = \mathbb E\left[e^{t(X-2)}|X\geq 2\right]=\dfrac{\mathbb E\left[e^{t(X-2)}\mathbf 1_{\{X\geq 2\}}\right]}{\mathbb P(X\geq 2)} = $$ $$=\dfrac{1}{e^{-2}}\int_0^\infty e^{t(x-2)}\mathbf 1_{\{x\geq 2\}}e^{-x}\,dx = \dfrac{1}{e^{-2}} \int_2^\infty e^{(t-1)x}e^{-2t}\,dx =\dfrac{1}{t-1}.$$
And your description of r.v. $X^*$ looks like the following: $$ X^*=\begin{cases}X-2, & X\geq 2\\ 0, & X<2\end{cases}=(x-2)\cdot\mathbf 1_{\{X\geq 2\}}+0\cdot\mathbf 1_{\{X< 2\}}. $$
In this case the distribution of $X^*$ is far from Exponential. It has an atom at $0$: $$\mathbb P(X^*=0)=\mathbb P(X<2)=1-e^{-2}$$ and therefore it is a mixture of pointmass at $0$ and the Exponential distribution. CDF for this $X^*$ is the following function discontinuous at zero: $$ F_{X^*}(t)=\mathbb P(X^*\leq t) =\begin{cases}0, & t<0\\ 1-e^{-(t+2)}, & t\geq 0\end{cases} $$