$\def\QQ{\mathbb{Q}}\def\ZZ{\mathbb{Z}}$In general, it is not true that $Gal(L/\QQ) \cong Gal(K/\QQ) \ltimes \mathrm{Cl}(K)$.
What is true
Suppose that we have a short exact sequence of groups
$$0 \to A \to H \to G \to 0$$
where $A$ is abelian. Since $A$ is normal, the group $H$ acts on $A$ by conjugation; since $A$ is abelian, this action factors through $H/A \cong G$. So, in any such setting, we get an action of $G$ on $A$.
In your notation, Galois theory gives us
$$0 \to \mathrm{Cl}(K) \cong Gal(L/K) \to Gal(L/\QQ) \to Gal(K/\QQ) \to 0.$$
It turns out that this action of $Gal(K/\QQ)$ is the same as the natural action of $Gal(K/\QQ)$ on the class group. I don't know a reference for this, but it's not hard to see that $\sigma \cdot \mathrm{Frob}_{\mathfrak{p}} \cdot \sigma^{-1} = \mathrm{Frob}_{\sigma \mathfrak{p}}$ for an unramified prime $\mathfrak{p}$ of $K$ and an element $\sigma \in G$. The result follows from this plus an extremely weak version of Cebataorv density (just enough to know that the values of Frobenius generate the Galois group).
However, this does not mean that the short exact sequence is semidirect and, as we will see below, it need not be.
A strategy for finding a counterexample
Suppose that we could construct a Galois extension $F/\QQ$, with Galois group $H$, and an abelian subgroup $A$ of $H$ such that
(1) The sequence $0 \to A \to H \to H/A \to 0$ did not split and
(2) For every prime $\mathfrak{p}$ of $F$ (including the infinite primes), the inertia group $I_{\mathfrak{p}}$ is disjoint from $A$.
I claim that taking $K$ to be the fixed field of $A$ will give a counterexample.
The extension $F/K$ is abelian (since $A$ is abelian) and unramified (by the condition on inertia groups). So we have $F \subseteq L$ and we have a commuting diagram
$$\begin{matrix}
0 &\to& \mathrm{Cl}(K) &\longrightarrow& Gal(L/\QQ) &\longrightarrow& Gal(K/\QQ) &\to & 0 \\
& & \downarrow & & \downarrow & & \| & & \\
0 &\to& A &\longrightarrow& Gal(F/\QQ) &\longrightarrow& Gal(K/\QQ) &\to & 0 \\
\end{matrix}$$
Suppose for the sake of contradiction that we had a map $Gal(K/\QQ) \to Gal(L/\QQ)$ splitting the top sequence. Then the composite map $Gal(K/\QQ) \to Gal(L/\QQ) \to Gal(F/\QQ)$ would split the bottom sequence, contradicting (1).
This makes it clear why counterexamples are hard to find. Since $\QQ$ has no unramified extensions, one of the $I_\mathfrak{p}$ must be nontrivial. So, just on the level of group theory, we need to find a non-semidirect short exact sequence $0 \to A \to H \to G \to 0$, with $A$ abelian and a nontrivial subgroup $I \subset H$ so that $I \cap A = \{ e \}$. If $I \to G$ is surjective, this implies that the sequence is semidirect after all. Just on the group theory level, we see that there are no examples with $G$ a cyclic group of prime order.
A counterexample
Let $\zeta$ be a primitive $85$-th root of unity. Recall that $$Gal(\QQ(\zeta)/\QQ) \cong (\ZZ/85)^{\ast} \cong (\ZZ/5)^{\ast} \times (\ZZ/17)^{\ast} \cong (\ZZ/4) \times (\ZZ/16).$$
We will always write elements of this group as ordered pairs in $(\ZZ/4) \times (\ZZ/16)$.
Let $F$ be the fixed field of $\{ 0 \} \times (4 \ZZ/16)$. So $Gal(F/\QQ) \cong (\ZZ/4) \times (\ZZ/4)$. The only ramified primes are $5$, $17$ and $\infty$ with inertia groups $\ZZ/4 \times \{0 \}$, $\{ 0 \} \times \ZZ/4$ and $(2,0)$ respectively. Let $A$ be the order $2$ subgroup generated by $(2,2)$. We see that $A \cap I_{\mathfrak{p}}$ is trivial for every $\mathfrak{p}$. Finally, we have $A \cong \ZZ/2$, $H \cong (\ZZ/4) \times (\ZZ/4)$ and $H/A \cong (\ZZ/4) \times (\ZZ/2)$, so the sequence is not semidirect.
I have not actually computed $\mathrm{Cl}(K)$ for this example.
One can build similar counterexamples whenever $H \cong (\ZZ/q^2)^2$ for some prime $q$, being a little careful about the prime at $\infty$ if $q=2$.
Smaller counterexample
I messed around a bit more, and concluded that there is nothing stopping me from taking $H$ to be the dihedral group of order $8$ and $A$ to be its (two-element) center. For example, let $F$ be the splitting field of $\QQ(\sqrt{1+8 \sqrt{-3}})$ (so $K = \QQ(\sqrt{-3}, \sqrt{193})$). If I haven't made any dumb errors, the only ramified primes are $3$, $193$ and $\infty$, and the inertia groups are all noncentral two-element subgroups. Even if I got this particular example wrong, I'm pretty sure there is no general obstacle to making an example like this.
Best Answer
Edit: The answer below is incorrect. While it is true that, via class field theory, we can recover the class group as a quotient of $G^{ab}$, the problem, as @ThePiper points out, is that this quotient is by $\widehat{\mathcal O}_K^\times$, which $G^{ab}$ knows nothing about.
Given the whole of $G$, we would be able to recover $\widehat{\mathcal O}_K^\times=\prod_{v}\widehat{\mathcal O}_{K_v}^\times$ via class field theory if we could recover the inertia groups $I_v$ from $G$: by local class field theory, $I_v\cong {\mathcal O}_{K_v}^\times$.
It is possible to recover the inertia groups from $G$. However, the fact that we can do so is a key part of the Neukirch-Uchida theorem.
The answer is yes. Let $G^{ab}$ denote the abelianisation of $G$ $-$ i.e. $G^{ab} = G/\overline{[G,G]}$. By global class field theory, we have a canonical isomorphism $$K^\times\backslash\mathbb A_K^{\times}/\overline{(K_\infty^\times)^0}\cong G^{ab}.$$
Here, $\mathbb A_K^\times$ are the ideles of $K$, and $\overline{(K_\infty^\times)^0}$ is the closure of the identity connected component of $(K\otimes_\mathbb Q\mathbb R)^\times$ viewed as a subgroup of $\mathbb A_K^\times$.
This isomorphism gives a concrete connection to the class group of $K$: the class group of $K$ is canonically isomorphic to $$K^\times\backslash\mathbb A_K^{\times}/\widehat{\mathcal O_K^\times} K_\infty^\times,$$ and is therefore a quotient of $G^{ab}$.
On the Galois side, this quotient of $G^{ab}$ cuts out a finite abelian extension of $K$ -- the Hilbert class field.