Convergence domain: $\{(z,w):|z|^2+|w|^2 < 1\}$

Hints:

1. Consider the series $\sum_{k=0}^\infty (z^2 + e^{ik} w^2)^k$.
2. If $|z^2| + |w^2| = \rho< 1$ then $|z^2 + e^{ik} w^2| \leq \rho$ for all $k$, and thus the series converges. If $|z^2| + |w^2| > 1$ then $|z^2 + e^{ik} w^2| > 1$ for infinitely many $k$, and the series diverges.

According to Lebl's book "Tasty Bits of Several Complex Variables", power series in several complex variables are considered for the absolute convergence. The reason behind that is non-existence of a canonical way of ordering terms in the power series of several variables.

So I will provide my answer according to following definitions:

Definition 1: A series $\sum c_{jk}z_1^jz_2^k$ is said to be absolutely convergent if the series $\sum \vert c_{jk}z_1^jz_2^k\vert$ convergent.

Definition 2: For a power series $\sum c_{jk}z_1^jz_2^k$, the domain of convergence is the interior of the subset of $\mathbb{C}^2$ where the power series is absolutely convergent.

Now answer to the question:

Start with a real power series of single variable $\sum a_nx^n$ whose radius of convergence is $1$ and replace $x^n$ with $(z_1^2e^{if_1(n)}+z_2^2e^{if_2(n)})^n$ where $f_1(n)$ and $f_2(n)$ any sequence of real numbers (or simply replace $x$ with $z_1^2+z_2^2$ and you get your answer.

But how? (I will show when $x$ is replaced with $z_1^2+z_2^2$ )

We have $$\sum_{n=0}^{\infty} a_n(z_1^2+z_2^2)^n=\sum_{n=0}^{\infty} \sum_{j=0}^n a_n \binom{n}{j}z_1^{2j}z_2^{2n-2j}\quad .$$ Now if you revisit definition 1 we have the following:

\begin{align} \sum_{n=0}^{\infty} \sum_{j=0}^n \vert a_n \binom{n}{j}z_1^{2j}z^{2n-2j}\vert=&\sum_{n=0}^{\infty} \sum_{j=0}^n \vert a_n\vert \binom{n}{j}\vert z_1 \vert^{2j}\vert z_2\vert^{2n-2j}\\=&\sum_{n=0}^{\infty} a_n(\vert z_1\vert^2+\vert z_2\vert^2)^n\quad . \end{align}

Now w.l.o.g. we can assume the real power series we started converges only when $\vert x \vert<1$. We can assume this because in definition 2 we only care about the interior. Then we see that $$\sum_{n=0}^{\infty} a_n(z_1^2+z_2^2)^n$$ converges absolutely if and only if $\vert z_1\vert^2+\vert z_2\vert^2<1$. So the series $$\sum_{n=0}^{\infty} a_n(z_1^2+z_2^2)^n$$ has desired domain of convergence.

For example: $$\sum_{n=1}^{\infty} (-1)^{n+1}\frac{(z_1^2+z_2^2)^n}{n}$$ is a specific answer to the question since we use Taylor's series of $\ln(1+x)$ here.