The claim is true, but you have definitely not proven it.
Case 1: you wrote $f(z-z) = f(z) - f(z)$. How do you know that? A priori all you know is $f(x+y) = f(x) + f(y)$, so here it becomes $f(z-z) = f(z+(-z)) = f(z) + f(-z)$. You still need to show that $f(-z) = -f(z)$, which is not part of what you assumed at the beginning.
To do case 1 properly, write $f(0) = f(0+0) = f(0) + f(0) = 2 f(0)$. So subtracting $f(0)$ from both sides you get $f(0) = 0$.
Case 2: You have the correct idea here. More formally you need to use induction: the base case is $f(1 \cdot z) = 1 \cdot f(z)$, which is obvious. For the induction step, assume that $f(kz) = kf(z)$ for a positive integer $k$, then $$\begin{align}
f((k+1)z) & = f(kz+z) \\
& = f(kz)+f(z) \quad \text{(hypothesis from the beginning)} \\
& = kf(z) + f(z) \quad \text{(induction hypothesis)}\\
& = (k+1)f(z).
\end{align}$$
Case 3: What you wrote is not really correct. In reality for $k$ a negative integer, $f(kz) = f((-z) + \dots + (-z))$ ($(-k)$ times), so by case 1 you find $f(-z) + \dots + f(-z) = -k f(-z)$. But then you are again stuck at $f(-z) = -f(z)$.
For a better proof, you can use case 1 and case 2 here: if $k$ is a negative integer, then $f(kz - kz) = f(kz) + f(-kz)$. But $f(kz-kz) = f(0) = 0$ by case 1, and since $(-k)$ is a positive integer, you can use case 2 to say $f((-k)z) = (-k)f(z)$, so $f(kz) - k f(z) = 0 \implies f(kz) = k f(z)$.
(Note: in particular for $k = -1$, this shows $f(-z) = -f(z)$. But again, you needed to prove that!)
This equation is known as Cauchy's functional equation. More generally you can show that $f(rz) = rf(z)$ is true for a rational number $r$, and if $f$ satisfies some additional property (continuity, monotonicity, measurability) then it's even true for every real number $r$. See this overarching question about the functional equation.
