Is $z \mapsto \int f(x,z) K_z(x) dx$ analytic if $K_z(x)$ is a distribution analytic in the paramter $z$?

Let $$K_z(x) \in \mathcal D'(\mathbb R)$$ be a distribution kernel depending analytically on $$z \in O \subset \mathbb C$$, i.e. for $$f \in \mathcal D(\mathbb R)$$ fixed, $$\int K_z(x) f(x) dx$$ is analytic for $$z$$ on an open set $$O$$.

Q: Is this still true if we multiply $$K_z(x)$$ with an analytic function $$a(x,z)$$ that is bounded on $$O$$? I.e., is $$z \mapsto \int f(x) a(x,z) K_z(x) dx$$ still analytic in $$O$$?

Attempt:
Showing that the integral satisfies the Cauchy-Riemann equations would be a way to prove that it is analytic, but is it possible to

1. Exchange integral and derivative $$\frac{d}{dz} \int f(x) a(x,z) K_z(x) dx = \int \frac{d}{dz} f(x) a(x,z) K_z(x) dx$$?
2. Argue that the Kernel $$K_z(x)$$ satisfies the CR-equations and not just $$\int f(x) K_z(x) dx$$?

Is there another approach?

Update:
While the answer by @reuns provides some inside, it is still missing the crucial point, i.e. exchanging two limit operations (in this case an integral and a derivative). To make this point more obvious, one can also consider $$(z,z') \mapsto \int f(x) a(x, z') K_z(x) dx$$ which is certainly analytic in $$z$$. It remains to show that it is analytic in $$z'$$. This boils down to the question whether distribution are linear in derivatives w.r.t. a paramter of the test function, i.e. $$\partial_z \int a(x,z) K(x) dx = \int \partial_z a(x,z) K(x) dx$$ This is something I suspect to be true, but am unsure about.

Take $$\phi \in D(\Bbb{R}),\int \phi = 1,\phi_n(x)=n \phi(nx)$$, let
$$H(y,z)= \int_\Bbb{R} f(x-y) a(x-y,z) K_z(x)dx$$ $$h_n(z) = \int_\Bbb{R} \phi_n(y)H(y,z) = \int_\Bbb{R}f(y) a(y,z) (\int_\Bbb{R} \phi_n(x-y) K_z(x)dx)dy$$ where $$y \to x-y$$ was substituted.
$$\int_\Bbb{R} \phi_n(x-y) K_z(x)dx$$ is continuous in $$y$$ and analytic in $$z$$ so by Morera's theorem, considering the sequence of Riemman sums $$h_{n,M}(z)=\frac1M\sum_{m=-\infty}^\infty f(m/M) a(m/M,z) \int_\Bbb{R} \phi_n(m/M-x) K_z(x)dx$$ $$h_n=\lim_{M\to \infty} h_{n,M}$$ is analytic.
$$H$$ is continuous in $$(y,z)$$ so $$h_n$$ converges locally uniformly as $$n\to \infty$$.
Whence by Morera's theorem the limit $$h_\infty(z)=H(0,z)=\int_\Bbb{R} f(x) a(x,z) K_z(x)dx$$ is analytic.
Here is a proof assuming $$K(z,x),a(x,z)$$ are analytic in $$z$$ and $$f(x) K(x,z)$$ is absolutely integrable in $$z,x$$. I am aware that the question wants for dsitribution kernel. But i thought this would be useful. Let $$g(z) = \int f(x) a(x,z) K(z,x) dx.$$ Now if $$\int_{\gamma} g(z) dz = 0$$ then by morera's theorem $$g(z)$$ is holomorphic. Now $$\int_{\gamma} g(z) dz = \int_{\gamma} \int f(x) a(x,z) K(z,x) dx dz.$$ Assuming the integral can be swapped, since $$a(x,z) K(z,x)$$ is analytic: $$\int_{\gamma} a(x,z) K(z,x) dz = 0$$. $$= \int f(x) \int_{\gamma} a(x,z) K(z,x) dz dx = 0.$$ So the proof is complete if we can swap the two iterated integrals. Since we assume $$f(x)K(z,x)$$ is absolutely integrable in $$x,z$$ with $$a(x,z)$$ bounded uniformly in $$x,z$$, we can swap the integrals by Fubini's theorem.