You are completely right regarding your question stated in the title: if $\phi\colon G\to G'$ is onto (i.e. surjective) then by definition the image $\phi(G)$ equals all of the codomain, that is $\phi(G)=G'$.
Anyway, I have to agree with André Armatowski. In my copy Lemma $8.15$ also reads as
Lemma $\textbf{8.15}$ Let $G$ and $G'$ be groups and let $\phi\colon G\to G'$ be a one-to-one function such that $\phi(xy)=\phi(x)\phi(y)$ for all $x,y\in G$. Then $\phi[G]$ is a subgroup of $G'$ and $\phi$ provides an isomorphism.
First off, the requirement that $\phi$ is one-to-one is unnecessary for the conclusion that $\phi[G]$ (I will adopt Fraleigh's notation) is a subgroup. To see this, consider a group homomorphism $\varphi\colon G\to H$ that is, a set-function $\varphi\colon G\to H$ such that $\varphi(xy)=\varphi(x)\varphi(y)~\forall x,y\in G$.
Claim. The image $\varphi[G]$ of $G$ under $\varphi$ is a subgroup of $H$.
Proof. As one can show directly from $\varphi$ being a group homomorphism, we have that $\varphi(e_G)=e_H$ and that $\varphi(g^{-1})=\varphi(g)^{-1}$. Therefore $\varphi[G]$ is non-empty. Now, let $\tilde x,\tilde y\in\varphi[G]$ then there exist $x,y\in G$ such that $\varphi(x)=\tilde x$ and $\varphi(y)=\tilde y$, respectively. Thus
$$\tilde x\tilde y^{-1}=\varphi(x)\varphi(y)^{-1}=\varphi(x)\varphi(y^{-1})=\varphi(xy^{-1})\in\varphi[G]$$
By the subgroup test, we conclude that $\varphi[G]$ is a subgroup of $H$.$~~~\square$
Lemma $8.15$ evolves around how we can obtain a trivial isomorphism from an injective function $\phi$ by restricting the codomain to the image of $\phi$. By definition, $\phi$ is surjective on its image and as $\phi$ is injective by assumption we obtain an isomorphism, i.e. a bijective group homomorphism. That's it.
