The problem gives the following piecewise function and asks for a proof of surjection.
$f: N \to N$ as defined by
$$f(x)= \begin{cases} x-1, & \text{if $x$ is odd,} \\ x+1, & \text{if $x$ is even.} \end{cases}$$
I know that 1 is not in the image of the function. But the nature of the piecewise function is causing me some confusion on how to formally prove this.
As is so often the case this depends upon whether your text considers $0$ to be a natural number or not.
As the book is claiming this function maps $\mathbb N \to \mathbb N$ and $f(1) = 1-1 = 0$, your text must consider $0$ to be a natural number.
So your claim there is no $f(n)=1$ is not true as $f(0) = 1$.
....
To prove this is surjective:
Show for any $m\in \mathbb N$ that there is an $n\in \mathbb N$ so that $f(n) = m$.
There are two cases:
1) $m$ is even so $m = 2k$; $m \ge 0$ and $k \ge 0$. So we must prove there is an $n$ so that $f(n)=2k$.
So
$\begin{cases}n\text{ is even}& f(n) = n+1 = 2k;&n=2k-1\text{ is odd; a contradiction}\\ n\text{ is odd}& f(n)=n-1=2k;&n=2k+1;\text{ is odd}&n\in \mathbb N\end{cases}$
We are done. For any even $2k$ we have $f(2k+1) =2k$.
2) $m$ is odd so $m = 2k+1$; $m \ge 1$ and $k \ge 0$. So we must prove there is an $n$ so that $f(n)=2k+1$.
So
$\begin{cases}n\text{ is odd}& f(n) = n-1 = 2k+1;&n=2k+2\text{ is even; a contradiction}\\ n\text{ is even}& f(n)=n+1=2k+1;&n=2k;\text{ is even}&n\in \mathbb N\end{cases}$
We are done. For any even $2k+1$ we have $f(2k) =2k+1$.
....
So for any $m\in\mathbb N$ there is an $n\in \mathbb N$ so that $f(n) =m$.