That someone told you wrong.
To prove that $\alpha: G \to G^{\prime}$ is a homomorphism between groups $(G,\,\cdot)$ and ($G^{\prime},\,*)$, you need to prove $\alpha(g_1\cdot g_2) = \alpha(g_1)*\alpha(g_2)$. You do not also need to prove $\alpha(g^{-1}) = \alpha(g)^{-1}$.
But it is true that for all homomorphisms $\alpha: G \to G^{\prime}$,
- $\alpha(e) = e^{\prime}$, where $e$ is the identity of $G$ and $e^{\prime}$ is the identity of $G^{\prime}$, and with this,
- it can also be proven that $\alpha(g^{-1}) = \alpha(g)^{-1}$ for $g\in G$.
- Furthermore, if $H$ is a subgroup of $G$, then $\alpha[G]$ is a subgroup of $G^{\prime}$, and
- if $K$ is a subgroup of $G^{\prime}$, then $\alpha^{-1}[K]$ is a subgroup of $G$.
In other words, a homomorphism $\alpha: G \to G^{\prime}$ maps identity to identity, inverses to inverses, and subgroups to subgroups.
But each of the above properties are necessarily implied by the property that defines a homomorphism: if you can show: $\alpha(g_1\cdot g_2) = \alpha(g_1)*\alpha(g_2)$ then you will have proven $\alpha: G \to G^{\prime}$ is a homomorphism between groups $(G,\,\cdot)$ and ($G^{\prime},\,*)$. The properties bulleted above then follow.