$\newcommand{\R}{\mathbb{R}}$$\newcommand{\R}{\mathbb{R}} \newcommand{\D}{\overset{\text{D}}=} \newcommand{\E}{\operatorname{\mathsf E}}$
The fundamental paper in this area is the one by Sharpe. By Theorem 1 in that paper, a full probability measure $\mu$ on $\R^d$ is (operator) stable if and only if for each natural $n$ there are a $d\times d$ real matrix $B_n$ and a vector $b_n\in\R^d$ such that $X_1+\cdots+X_n$ equals $B_nX+b_n$ in distribution, where $X,X_1,X_2,\ldots$ are independent random vectors in $\R^d$ each with distribution $\mu$. (A measure $\mu$ on $\R^d$ is called full if the linear span of its support is $\R^d$.)
If a random vector $X$ in $\R^d$ has a stable distribution, we may say that $X$ itself is stable. Examples of stable random vectors include random vectors with independent stable coordinates. Also, if a a random vector $X$ in $\R^d$ is stable, then any nonsingular affine transformation $BX+b$ of $X$ is also obviously stable (here $B$ is any nonsingular $d\times d$ real matrix and $b\in\R^n$). In particular, any Gaussian distribution on $\R^d$ is stable.
Added: In particular, it follows that the understanding of the notion of operator-stable distributions highlighted in the OP's answer is incorrect, even with the additional assumption that $X_1$ and $X_2$ be independent. Indeed, consider, e.g., the case when $X=\begin{bmatrix}\xi\\ \eta\end{bmatrix}$, where $\xi$ and $\eta$ are independent real-valued random variables with the standard Cauchy and Gaussian distributions, respectively. Then the distribution of $X$ is operator-stable. Let $X_1=\begin{bmatrix}\xi_1\\ \eta_1\end{bmatrix}$ and $X_2=\begin{bmatrix}\xi_2\\ \eta_2\end{bmatrix}$ be independent copies of $X$. Let $A:=\begin{bmatrix}1&0\\0&1\end{bmatrix}$ and $B:=\begin{bmatrix}0&1\\1&0\end{bmatrix}$, and suppose that \begin{equation} Y:=AX_1+BX_2\D CX+d=:Z \tag{1} \end{equation} for some real matrix $C=\begin{bmatrix}c_{11}&c_{12}\\ c_{21}&c_{22}\end{bmatrix}$ and a vector $d$. Then $d=0$, since the distributions of the random vectors $Y$ and $CX$ are symmetric about $0$. We have $Y=\begin{bmatrix}\xi_1+\eta_2\\ \eta_1+\xi_2\end{bmatrix}$, and so, for any $t=\begin{bmatrix}r\\ s\end{bmatrix}\in\R^{2\times1}$, \begin{equation} \E e^{it\cdot Y}=\exp\{-|r|-|s|-r^2/2-s^2/2\}, \end{equation} whereas \begin{equation} \E e^{it\cdot Z}=\exp\{-|rc_{11}+sc_{21}|-(rc_{21}+sc_{22})^2/2\}, \end{equation} so that \begin{equation} |r|+|s|+(r^2+s^2)/2=|rc_{11}+sc_{21}|+(rc_{21}+sc_{22})^2/2 \end{equation} for all real $r,s$. Letting here $r\to\infty$ and $s=0$, we have $c_{21}=\pm1$. Similarly, $c_{22}=\pm1$. On the other hand, letting $r=s\to\infty$, we have $(c_{21}+c_{22})^2/2=1$, which contradicts the conditions $c_{21}=\pm1$ and $c_{22}=\pm1$. This contradiction shows that (1) cannot hold here for any $C,d$.