Terminology: A subvariety $V$ of an abelian variety $A/\mathbb{C}$ is called a product if there are integral closed subvarieties $U,W\subset A$ such that $\dim U,\dim W>0$ and the sum morphism $U\times W\to A$ is a closed immersion with image $V$. (Def., p.3 of Arithmetic finiteness of very irregular varieties , Krämer-Maculan, https://arxiv.org/pdf/2310.08485v1.pdf .)
A result: Let $X/\mathbb{C}$ be a smooth integral projective variety with a base point $x\in X(\mathbb{C})$. If the Albanese morphism $\mathrm{alb}_x:X\to \mathrm{Alb}(X)$ is a closed immersion with ample normal bundle, and $X$ is the product of $X_1,X_2\subset \mathrm{Alb}(X)$, then $X=\mathrm{Alb}(X)$. (This fact, told by Professor Maculan, is related to Def., p.1 in https://arxiv.org/pdf/2310.08485v1.pdf .)
Proof: In fact, by translating, one may assume that $0\in X_1,X_2$. Then $\mathrm{Alb}(X)=\mathrm{Alb}(X_1)\times \mathrm{Alb}(X_2)$. By Prop 1.3 of Fulton-Hansen and Barth-Lefschetz theorems for subvarieties of abelianvarieties (O. Debarre), $X$ is geometrically non-degenerate in $\mathrm{Alb}(X)$. The image of $X$ under the projection $\mathrm{Alb}(X)\to\mathrm{Alb}(X_1)$ coincides with the image of $X_1$. Since $\dim X_1=\dim X-\dim X_2<\dim X$, by p.188 in Debarre's article., $X_1=\mathrm{Alb}(X_1)$. Similarly, $X_2=\mathrm{Alb}(X_2)$ and $X=\mathrm{Alb}(X)$.
Every integral closed subvariety of a simple abelian variety has ample normal bundle. This leads me to the following question.
Question: Is there an example of two integral closed subvarieties $X,Y$ of a simple abelian variety $A/\mathbb{C}$, such that $\dim X,\dim Y>0$ and the sum morphism $X\times Y\to A$ is a closed immersion?
