This is stated as a corollary from the Homotopy Axiom (Rotman) which states the if $f \sim g$ are homotopic maps, then $H_n(f) = H_n(g)$ for $n \geq 0$. (going to drop the $n$)
Because $X$ and $Y$ have the same homotopy type, there exist $f: X \to Y$ and $g: Y \to X$ such that $f \circ g \sim id_Y$ and $g \circ f \sim id_X$. By the Axiom, $$id_{H(Y)} = H(id_Y) = H(f\circ g)$$ $$H(g \circ f) = H(id_X) = id_{H(X)}.$$
So $H(f) \circ H(g) = id_{H(Y)}$ and $H(g) \circ H(f) = id_{H(X)}$
Aren't these just "left" and "right" inverses? How do we draw the isomorphism?
There is nothing left to be shown. You have a group homomorphism $H(f)$ which has an inverse, namely $H(g)$, and that is the definition of an isomorphism (in any category).