PROOF:
Let be $\begin{aligned}[t] \rho\colon G &\longrightarrow \operatorname{Sym}(G)\\ g &\longmapsto g^{\rho}\colon G \rightarrow G\\ & \qquad \;x\longmapsto xg \end{aligned}$
Thus $xg=x \iff g=\text{Identity} \implies \rho$ is faithful. (i.e., $\operatorname{kernel}(\rho)=\{1\}$).
${\color{red}{\text{My question is: Where is the isomorphism?}}}$
${\color{red}{\text{I didn't understand how the isomorphism theorem was applied here.}}}$
I'd go to the conclusion by steps.
Initially we have just the following setting:
\begin{alignat*}{2} \rho:G &\longrightarrow& X(G) \\ a&\longmapsto& \rho_a:G &\longrightarrow G \\ &&g&\longmapsto\rho_a(g):=ag \\ \tag 1 \end{alignat*}
where $X(G)$ is the set of the maps of $G$ in itself (we still don't know whether $\rho_a$ is a bijection for every $a \in G$). Now, $\forall a \in G, \rho_a(g)=\rho_a(h) \Rightarrow ag=ah \Rightarrow g=h$, so $\rho_a$ is injective for every $a \in G$. Besides, $\forall a,g \in G, g=\rho_a(a^{-1}g)$, so $\rho_a$ is also surjective for every $a \in G$. Conclusion: $\forall a \in G, \rho_a \in S_G$, and $(1)$ can indeed be refined into yours:
\begin{alignat*}{2} \rho:G &\longrightarrow& S_G \\ a&\longmapsto& \rho_a:G &\longrightarrow G \\ &&g&\longmapsto\rho_a(g):=ag \\ \tag 2 \end{alignat*}
Furthermore, $a,b \in G \Rightarrow ab \in G \Rightarrow \rho_{ab} \in S_G$, so to state anything about $\rho_{ab}$, we have to let it work as map (bijection) on $G$:
$$\forall g \in G, \rho_{ab}(g)=(ab)g=a(bg)=\rho_a(\rho_b(g))=(\rho_a\rho_b)(g) \Rightarrow \rho_{ab}=\rho_a\rho_b \tag 3$$
and thence $\rho$ is a homomorphism.
Finally, we are left to prove that $\rho$ itself is injective:
$$\rho_a=\rho_b \Rightarrow \rho_a(g)=\rho_b(g), \space\forall g \in G \Rightarrow ag=bg,\space\forall g \in G \Rightarrow a=b \tag 4$$
and indeed $G$ embeds in $S_G$ (by left multiplication) or, equivalently, $\rho$ is an isomorphism from $G$ to $\rho(G)\le S_G$. This is Cayley's Theorem.
[ Alternatively to $(4)$: $$a\in \operatorname{ker}\rho \Rightarrow \rho_a=\iota_G \Rightarrow \rho_a(g)=\iota_G(g), \forall g\in G \Rightarrow ag=g, \forall g \in G \Rightarrow a=e$$
whence $\operatorname{ker}\rho = \{e\}$. ]