In order to prove that $G$ acts on itself by multiplication, we have to verify that $(gh)\cdot x = g\cdot (h\cdot x)$ and $e\cdot x = x$ for all $g, h, x\in G$. These properties follow directly from the associativity of $\cdot$ and the fact that $e$ is the neutral element of $G$. Since $G$ is a finite group, left multiplication just permutes the group elements, i.e. $J(g)\in S(G)$ for all $g\in G$. Obviously, $J\colon G\to S(G)$ is a group homomorphism because $J(g_1g_2)(h) = (g_1g_2)h = g_1(g_2h) = J(g_1)J(g_2)(h)$ for all $g_1,g_2,h\in G$. Furthermore, $J$ is injective: Its kernel consists of all $g\in G$ that satisfy $gh = h$ for all $h\in G$ (in particular $ge = e$), and this is true if and only if $g=e$. Thus, $G\cong G/\{e\} = G/\ker J$. By the first isomorphism theorem, we also have $G/\ker J\cong \im J$, so combining these results, we see that $G$ is isomorphic to $\im J$, which is a subgroup of $S(G)$ (and thus isomorphic to a subgroup of $S(n)$, $n=|G|$).