$\def\QQ{\mathbb{Q}}\def\ZZ{\mathbb{Z}}$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.