Let $a$ and $b$ are integers.
If $a \ne b$, then $ab \ne 1$.
Proving the contrapositive: If $ab = 1$ then $a = b$. If $ab = 1$, then there are two possibilities. $a = b = 1$ or $a = b = -1$. No other choice of $a$ and $b$ can make $ab = 1$. In both these cases, $a = b$.
Is the proof correct? Can I have a direct proof?
Your proof is correct, although you are a bit fast on the part "the only options are $a=b=1$ and $a=b=-1$.
A direct proof can be as follows:
Assume $ab=1$. Then $a\neq 0$ and $b\neq 0$.
If $a>1$ or $a<-1$ then $a$ has a prime factor $p$, therefore $p\mid ab$ which contradicts $ab=1$.
Therefore $a\in\{-1,1\}$ and by symmetry $b\in\{-1,1\}$ as well. Now $a$ and $b$ have to be equal otherwise $ab=-1$.
Let $a, b \in \mathbb{Z}$ with $a \neq b$. If one of those integers is positive and the other negative, the product is negative. If either $a=0$ or $b=0$, the product is zero.
So we only have to look at the case of equal signs, let's start with $a>0$ and $b>0$. Since $a \neq b$, one of the integers needs to be larger than $1$. Without loss of generality, suppose $a>1$, then $ab > b \geq 1$, thus $ab \neq 1$.
You can do the case that both are negative in an analogous way.
Assume that $a,b \in \mathbb{Z}$.
If $a$ or $b$ is zero, the implication holds. If $a,b \not = 0$ and assume without loss of generality that $a<0$ and $b>0$. In this case we have that $ab<0\not =1$. Lastly if $a,b\not = 0$ and wlg. $a,b >0$ then $a,b \in \{1,2,3,...\}$ and we see that $ab=1$ iff $a=b=1$ which is a contradiction.
