You need compact convergence of the series, that is, convergence on compact sets. But we also need some more preliminaries:
First, I'm not going to talk about the indefinite integral, but just about antiderivatives, simply because we will need definite integrals down the line, and I think it will be less confusing if we reserve the integral sign for the definite ones. Second, the expression $\sum\int a_nz^n\mathrm dz$ is a bit icky, since the integrals all come with an arbitrary additive constant attached, which can make the sum diverge. We will have to specify which antiderivative we are choosing specifically. Third, we will only be considering connected domains, for simplicity's sake. You will soon see why. Fourth, interchanging sums and integrals or sums and antiderivatives is really about interchanging limits and integrals/antiderivatives, since infinite sums are just sequences/their limits written down in a particular way. So the real question is: If $f_n\to f$, does $F_n\to F$, where $F_n,F$ are suitably chosen antiderivatives of $f_n$ and $f$? And in what manner do they converge?
Now with these remarks out of the way, we can say the following (which is about sequences of functions, but series are sequences, so you can apply it to series exactly the same way):
Let $D\subseteq\mathbb C$ be a connected domain. Let $f_n:D\to\mathbb C$ be holomorphic for all $n\in\mathbb N$, and let $f_n$ converge to $f:D\to\mathbb C$ uniformly on compact subsets of $D$. Also, let all $f_n$ have an antiderivative on $D$.
Then the function
$$F:D\to \mathbb C,~z\mapsto\int_{z_0}^z f(w)\mathrm dw,$$
where the integral goes along any arbitrary path from a fixed $z_0\in D$ to $z$, is well defined and an antiderivative of $f$. We also have $F_n\to F$ uniformly on compact sets, where $F_n$ is an analogously defined antiderivative of $f_n$.
Proof:
First note that because $f_n\to f$ uniformly on compact subsets, $f$ is holomorphic. Also, since $D$ is connected and open, it is also path connected, so
a path from $z_0$ to $z$ is guaranteed to exist. And since the functions $f_n$ admit an antiderivative, the integral
$$F_n(z)=\int_{z_0}^z f_n(w)\mathrm dw$$
does not depend on the path and is thus well-defined. Due to uniform convergence of the integrand to $f$ on the arbitrarily chosen path (it's compact), the integral converges to
$F(z)=\int_{z_0}^z f(w)\mathrm dw,$ which thus also doesn't depend on the path and is then well-defined.
If we can show that the convergence is uniform on compact subsets, then $F$ is holomorphic and $F_n'\to F'$ uniformly on compact subsets, too. But since $F_n'=f_n$ and $f_n\to f$, we will then have $F'=f$, so $F$ is an antiderivative of $f$. So we show this uniform convergence on compact subsets:
Note that any compact subset of $D$ can be covered by a finite number of compact discs. And if a function converges uniformly on a finite number of sets, then it also converges uniformly on their union. So it is sufficient to show uniform convergence on compact discs. Let $\overline{U_r}(z_\ast)\subset D$ be such a disc with radius $r$ centered at $z_\ast$. On this disc we have
$$\begin{align}
\vert F(z)-F_n(z)\vert&=\left\vert\int_{z_0}^z f(w)-f_n(w)\mathrm dw\right\vert\\
&=\left\vert\int_{z_0}^{z_\ast} f(w)-f_n(w)\mathrm dw+\int_{z_\ast}^z f(w)-f_n(w)\mathrm dw\right\vert\\
&=\left\vert F(z_\ast)-F_n(z_\ast)+\int_{z_\ast}^z f(w)-f_n(w)\mathrm dw\right\vert\\
&\leq\vert F(z_\ast)-F_n(z_\ast)\vert + \left\vert\int_{z_\ast}^z f(w)-f_n(w)\mathrm dw\right\vert\\
&\leq \vert F(z_\ast)-F_n(z_\ast)\vert + r\sup_{w\in\overline{U_r(z_\ast)}}\vert f(w)-f_n(w)\vert.
\end{align}$$
The last estimation gives a bound that doesn't depend on $z$ and goes to $0$ (the first term because $F_n\to F$ pointwise, the second because $f_n\to f$ uniformly on the compact disc). So $F_n\to F$ uniformly on the disc, and by the argument above, on any other compact set as well.
