The answer is positive, and the abelian case can be handled even over 'Hilbertian fields'.

By the Schur-Zassenhaus Theorem the exact sequence splits, so we get a so called split embedding problem. If the kernel is abelian, the embedding problem is regularly solvable. That is, it is (properly) solvable over the function field $K(T)$. Since $K$ is Hilbertian (i.e. satisfies Hilbert's irreducibility theorem), the solution over $K(T)$ can be specialized to a solution over $K$ in infinitely many distinct ways. In fact, the infinitely many specializations can be taken linearly disjoint over the fixed field of $\mathrm{Ker}(\varphi)$ (i.e. $L$).

For more details see the book 'Field Arithmetic' by Fried and Jarden.

More generally, consider the embedding problem where $G$ is replaced by its $n$-th fibered power over $\varphi$. The solvability of this embedding problem (which follows from the result stated by the OP) implies that the original embedding problem has $n$ distinct solutions for each positive integer $n$, as required.

The $n$-th fibered power $G^n_{\varphi}$ of $G$ over $\varphi$ is the subgroup of all $(g_1, \dots,g_n) \in G^n$ such that $\varphi(g_i) = \varphi(g_1)$ for all $1 \leq i \leq n$. It fits into the exact sequence

$$1\to \ker(\varphi|_G)^n \to G^n_{\varphi} \overset{\varphi}{\to} \text{Gal}(L/K)\to 1.$$

Even more generally, it was conjectured that the answer to this question is positive also in the case that the kernel is not solvable. Specifically, Debes and Deschamps conjectured that every finite split embedding problem is solvable over a number field (so in particular, the inverse Galois problem has an affirmative answer).