I need to prove that every element of the fundamental group has an inverse.
First we define a map $\phi:I\to I$ homotopic to $\operatorname{Id}_I$. If $\phi$ is the constant zero function isn't it true that $f\circ \phi \simeq f$? So if we denote the inverse of $f$ by $f'$ isn't it true that $(f\ast f')\circ \phi \simeq f\ast f'$ and $(f\ast f')\circ \phi = e$, which means $e\simeq f\ast f'$? ($\ast$ is the product of paths and $e$ is the identity element)
But the problem is that I didn't use any property of the inverse element, so something is wrong. How can I fix it?
Your argument is incorrect. $f\circ\phi$ is not in general homotopic to $f$- it's homotopic to the constant map taking the value $f(0)$.
To prove that every element of the fundamental group has an inverse, think about some examples of elements of the fundamental group and their inverses. You should see a pattern jumping out that suggests a natural course for your proof.