I've upvoted vadim123's bijective answer (which I think is the best way of seeing this), but if you want to do it via your original approach of generating functions, note that the second one you wrote is not correct.
A composition where the first part is $1$, or in other words something that has the form "$1$" followed by a sequence of positive integers, can be denoted (following the notation of the book Analytic Combinatorics) as $\mathcal{Z} \times \operatorname{S\scriptsize EQ}(\mathcal{I})$ where $\mathcal{I}$ denotes the class of positive integers $\{1, 2, 3, \dots\}$ (with the "size" of an integer being that integer itself). In other words $\mathcal{I} = \operatorname{S\scriptsize EQ}_{\ge 1}(\mathcal{Z})$ and it has generating function $\frac{z}{1-z}$, which means that the GF for compositions with first part $1$ is
$$z \frac{1}{1-\frac{z}{1-z}} = \frac{z(1-z)}{1-2z},$$
as you got.
The class of compositions where the first part is greater than $1$, and therefore at least $2$, is
$$\operatorname{S\scriptsize EQ}_{\ge 2}(\mathcal{Z}) \times \operatorname{S\scriptsize EQ}(\mathcal{I}) \\
= \operatorname{S\scriptsize EQ}_{\ge 2}(\mathcal{Z}) \times \operatorname{S\scriptsize EQ}(\operatorname{S\scriptsize EQ}_{\ge 1}(\mathcal{Z}))
$$
and therefore has generating function
$$\frac{z^2}{1-z} \frac{1}{1-\frac{z}{1-z}} = \frac{z^2}{1-2z},
$$
different from what you got.
These are different expressions, but if we denote $T_n = [z^n]\frac{1}{1-2z}$, then the coefficient of $z^n$ in the first GF is $T_{n-1} - T_{n-2}$, and in the second GF is $T_{n-2}$, which happen to both be equal to $2^{n-2}$ as $T_n = 2^n$ for all $n$.