If a function has an inverse then it is bijective?

If you have $f: X \to Y$, $g : Y \to X$ with $g(f(x)) = x$ for all $x \in X$, it is still possible for $f$ to not be bijective. However, $f$ will be a bijection onto its image; i.e., $f$ is a bijection from $X$ to $f(X)$. In particular, $f$ is injective.

If you additionally require that $f(g(y)) = y$ for all $y \in Y$ (i.e. $g$ is a "two-sided inverse"), then $f$ is a bijection from $X$ to $Y$.

Not necessarily. Simply, the fact that it has an inverse does not imply that it is surjective, only that it is injective in its domain.

Take for example the functions $f(x)=1/x^n$ where $n$ is any real number. As $x$ approaches infinity, $f(x)$ will approach $0$, however, it never reaches $0$, therefore, though the function is inyective, and has an inverse, it is not surjective, and therefore not bijective.