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.