Since $p|ab$, then $\exists \alpha \in\ \mathbb{Z}$ such that:
$$\alpha p=ab$$
By dividing both sides by $a$, we obtain:
$$(\alpha \div b)p=a \Rightarrow\ p|a \ \ ( \text{Similarly to prove that} \ \ p|b)$$
Is this proof comprehensive? Does it fulfill what is required?
It's not complete because it's possible that $b$ might not divide $\alpha$. For example, consider $$p | \alpha p$$ with $p \nmid \alpha$.
Then indeed $\alpha p = \alpha p$ but $\alpha \div p$ is not an integer number.