Your idea almost works. Make no change when $k \ge 0$. When $k<0$, send $k$ to $-2+\frac{1}{|k|}$.
Comment: Before we get to formulas, let's think about what the set $A$ looks like. It consists of
$2+\frac{1}{1}$, $2+\frac{1}{2}$, $2+\frac{1}{3}$, and so on, together with
$-2+\frac{1}{1}$, $-2+\frac{1}{2}$, $-2+\frac{1}{3}$, and so on.
It seems reasonable to send $0$ to $2+\frac{1}{1}$, to send $1$ to $2+\frac{1}{2}$, to send $2$ to $2+\frac{1}{3}$, and so on. The definition of $f(k)$ for $k \ge 0$ is then clear.
Then it seems natural to send $-1$ to $-2+\frac{1}{1}$, to send $-2$ to $-2+\frac{1}{2}$, to send $-3$ to $-2+\frac{1}{3}$, and so on. The definition of $f(k)$ for $k < 0$ is then clear. (Instead of writing $2-\frac{1}{|k|}$ we could write $f(k)=-2-\frac{1}{k}$. But I feel more comfortable with positive numbers.)
The fact that $f$ is bijective is intuitively absolutely obvious, and (almost) does not require proof. However, if proof is asked for, it is easy to write down.