Theorem: Let $L:K$ be a finite extension inside $\mathbb{C}$, such that the fixed field of $G = Gal(L:K)$ is $K$. Then $L:K$ is normal.
Proof: Let $m$ be any irreducible polynomial over $K$ with a zero $x$ in $L$. Since the extension is finite, so is the order of $G$. Consider now $G(x) =$ {$g(x) : g \in G$}.
We show that any $g' \in G$ permutes the elements in $G(x)$.
Write $G(x)=$ {$x_1, ..., x_n$}. Let $g' \in G$ and without loss of generality put $g'(x) = g_1(x) = x_1$. There exist $g_2, ..., g_n \in G$ such that $g_i(x) = x_i$. So that we have {$g_1, ..., g_n$}$\subseteq G$ with $G(x) =$ {$g_1(x), ..., g_n(x)$}. Define an operation on $G(x)$ by
$$x_i \cdot x_j = g_i(x) \cdot g_j(x) = g_ig_j(x)$$
$G(x)$ forms a group under such operation:
Closure: $x_i \cdot x_j = g_ig_j(x) \in G(x)$
Identity: $x \in G(x)$, thus $g_r(x) = x$ for some $r$. Therefore for any $x_i \in G(x)$, we have $x_i \cdot x_r = g_ig_r(x) = g_i(x) = x_i$. Similarly $x_r \cdot x_i = x_i$ for any $x_i \in G(x)$.
Inverses: for all $x_i \in G(x)$, we have $g_i^{-1}(x) = g_j(x) \in G(x)$. So that $x_i \cdot x_j = g_ig_j(x) = x$. Similarly $x_j \cdot x_i = x$.
Associativity: $(x_a \cdot x_b) \cdot x_c = g_ag_b(x) \cdot g_c(x) = g_ag_bg_c(x) = g_a(x) \cdot g_bg_c(x) = x_a \cdot (x_b \cdot x_c)$.
By standard Group Theory $g' = g_1$ permutes $G(x)$.
Now let $p(t) = (t-x_1)...(t-x_n)$. Since any element $g \in G$ permutes {$x_1, ..., x_n$}, we have $g(p(t)) = p(t)$ for all $g \in G$, and thus $p(t) \in G_0[t]$ where $G_0$ is the fixed field of $G$. But such field is $K$, therefore $p(t) \in K[t]$. Now $m$ divides $p$ in $K[t]$, so that $p = qm$. All the zeroes of $m$ are then in $G(x)$, but $G(x) \subseteq L$, thus $m$ splits in $L$.
Is the proof above correct?
It's not generally true that the selected subset of $G$, $\{g_1,\ldots,g_n\}$ forms a group. You're trying to argue that a transitive group action induces a group structure on a set, which is not true in general. For example, how does $S_n$ acting on $\{1,\ldots,n\}$ induce a group structure on $\{1,\ldots,n\}$?
Thus your proofs of existence of identity and inverses fail in general, because it's not clear (indeed, it isn't true generally) that $g_ig_j$ will equal the element of $G$ that you've selected that sends $x$ to $g_ig_j(x)$. Thus $x_r \cdot x_i = (g_rg_i)x\ne g_ix=x_i$ generally for identity. There is a similar problem in your proof of the existence of inverses. We have $g_i^{-1}x=x_j$ for some $j$. That gets you $x_i\cdot x_j = g_i(g_j x) = g_i(g_i^{-1}x)=x=x_r$. However $x_jx_i = g_jg_i x$, and all you know is that $g_ig_jx=x$, not that $g_jg_ix=x$.
That said, it should be clear that any element of $G$ permutes the elements of $G(x)$ without all of what you wrote, since for any $gx\in G(x)$ and $g'\in G$, $g'(gx)=(g'g)x\in G(x)$. The rest looks correct to me.
Edit
While what you want to do doesn't work in general, there is a special case in which it does. If $G$ acts transitively and freely on a set $X$, then you can put a group structure on $X$ making it isomorphic to $G$ by choosing a special element $x\in X$ in essentially the way you are trying to do here. See here for more info.