In two papers dated 2014 and 2016 (i.e., after 6 years the question was posted), this question was completely solved in genus 2 case by E. Kani. As Francesco Polizzi mentioned, this question was partially solved in 1965 by Hayashida and Nishi. They only worked on the abelian varieties $A$ over $\C$ with the product of two $\CM$ elliptic curves by the maximal order. So there are two things to remove the assumptions: 1) $A$ may be $E_1\times E_2$ with their endomorphism rings are not necessarily isomorphic. 2) Their endomorphism ring may be any order, i.e., not necessarily the maximal order.

By using the refined Humbert invariant which is an integral quadratic form, Kani translated this geometric problem into an arithmetic problem and removed these two assumptions.

Here is the statement for the $\CM$ case from [JT]:=Jacobians isomorphic to a product of two elliptic curves and ternary quadratic forms, 2014.:

Let $E_1\sim E_2$ be two isogenous $\CM$ elliptic curves over an algebraically closed field $K$ with its characteristic is zero. Then there is no genus 2 curve on $E_1\times E_2$ if and only if the degree map on $\Hom(E_1,E_2)$ is equivalent to one of the 15 forms $f(x,y)=ax^2+bxy+cy ^2$ whose coefficients $(a,b,c)$ are in the following list: $\mathcal{S}=\{k(1,1,1):k=1,2,4,6,10\}\cup\{k(1,0,1):k\in 1,2,6\}\cup\{(1,1,2),(1,1,4)\}\cup\{2(1,1,c):c=3,9\}\cup\{2(1,0,c):c=2,5\}\cup\{2(2,0,3)\}.$

The statement for the nonCM case can be found in Theorem 1 of [JT]. Its proof can be found in the article of The moduli spaces of Jacobians isomorphic to a product of two elliptic curves, 2016.

Some notes: The assumption of the characteristic 0 can be removed. This was also done by Ibukiyama/Katsura/Oort.

The exact list is only known under the GRH in the nonCM case. But Kani's solution doesn't depend on the GRH.

In two papers dated 2014 and 2016 (i.e., after 6 years the question was posted), this question was completely solved in genus 2 case by E. Kani. As Francesco Polizzi mentioned, this question was partially solved in 1965 by Hayashida and Nishi. They only worked on the abelian varieties $A$ over $\mathbb{C}$ with the product of two $CM$ elliptic curves by the maximal order. So there are two things to remove the assumptions: 1) $A$ may be $E_1\times E_2$ with their endomorphism rings are not necessarily isomorphic. 2) Their endomorphism ring may be any order, i.e., not necessarily the maximal order.

By using the refined Humbert invariant which is an integral quadratic form, Kani translated this geometric problem into an arithmetic problem and removed these two assumptions.

Here is the statement for the $CM$ case from [JT]:=Jacobians isomorphic to a product of two elliptic curves and ternary quadratic forms, 2014.:

Let $E_1\sim E_2$ be two isogenous $CM$ elliptic curves over an algebraically closed field $K$ with its characteristic is zero. Then there is no genus 2 curve on $E_1\times E_2$ if and only if the degree map on $Hom(E_1,E_2)$ is equivalent to one of the 15 forms $f(x,y)=ax^2+bxy+cy ^2$ whose coefficients $(a,b,c)$ are in the following list: $\mathcal{S}=\{k(1,1,1):k=1,2,4,6,10\}\cup\{k(1,0,1):k\in 1,2,6\}\cup\{(1,1,2),(1,1,4)\}\cup\{2(1,1,c):c=3,9\}\cup\{2(1,0,c):c=2,5\}\cup\{2(2,0,3)\}.$

The statement for the nonCM case can be found in Theorem 1 of [JT]. Its proof can be found in the article of The moduli spaces of Jacobians isomorphic to a product of two elliptic curves, 2016.

Some notes: The assumption of the characteristic 0 can be removed. This was also done by Ibukiyama/Katsura/Oort.

The exact list is only known under the GRH in the nonCM case. But Kani's solution doesn't depend on the GRH.

In two papers dated 2014 and 2016 (i.e., after 6 years the question was posted), this question was completely solved in genus 2 case by E. Kani. As Francesco Polizzi mentioned, this question was partially solved in 1965 by Hayashida and Nishi. They only worked on the abelian varieties $A$ over $\C$ with the product of two $\CM$ elliptic curves by the maximal order. So there are two things to remove the assumptions: 1) $A$ may be $E_1\times E_2$ with their endomorphism rings are not necessarily isomorphic. 2) Their endomorphism ring may be any order, i.e., not necessarily the maximal order.

By using the refined Humbert invariant which is an integral quadratic form, Kani translated this geometric problem into an arithmetic problem and removed these two assumptions.

Here is the statement for the $\CM$ case from [JT]:=Jacobians isomorphic to a product of two elliptic curves and ternary quadratic forms, 2014.: Let $E_1\sim E_2$ be two isogenous $\CM$ elliptic curves over an algebraically closed field $K$ with its characteristic is zero. Then there is no genus 2 curve on $E_1\times E_2$ if and only if the degree map on $\Hom(E_1,E_2)$ is equivalent to one of the 15 forms $f(x,y)=ax^2+bxy+cy ^2$ whose coefficients $(a,b,c)$ are in the following list: $\mathcal{S}=\{k(1,1,1):k=1,2,4,6,10\}\cup\{k(1,0,1):k\in 1,2,6\}\cup\{(1,1,2),(1,1,4)\}\cup\{2(1,1,c):c=3,9\}\cup\{2(1,0,c):c=2,5\}\cup\{2(2,0,3)\}.$

The statement for the nonCM case can be found in Theorem 1 of [JT]. Its proof can be found in the article of The moduli spaces of Jacobians isomorphic to a product of two elliptic curves, 2016.

In two papers dated 2014 and 2016 (i.e., after 6 years the question was posted), this question was completely solved in genus 2 case by E. Kani. As Francesco Polizzi mentioned, this question was partially solved in 1965 by Hayashida and Nishi. They only worked on the abelian varieties $A$ over $\C$ with the product of two $\CM$ elliptic curves by the maximal order. So there are two things to remove the assumptions: 1) $A$ may be $E_1\times E_2$ with their endomorphism rings are not necessarily isomorphic. 2) Their endomorphism ring may be any order, i.e., not necessarily the maximal order.

By using the refined Humbert invariant which is an integral quadratic form, Kani translated this geometric problem into an arithmetic problem and removed these two assumptions.

Here is the statement for the $\CM$ case from [JT]:=Jacobians isomorphic to a product of two elliptic curves and ternary quadratic forms, 2014.:

Let $E_1\sim E_2$ be two isogenous $\CM$ elliptic curves over an algebraically closed field $K$ with its characteristic is zero. Then there is no genus 2 curve on $E_1\times E_2$ if and only if the degree map on $\Hom(E_1,E_2)$ is equivalent to one of the 15 forms $f(x,y)=ax^2+bxy+cy ^2$ whose coefficients $(a,b,c)$ are in the following list: $\mathcal{S}=\{k(1,1,1):k=1,2,4,6,10\}\cup\{k(1,0,1):k\in 1,2,6\}\cup\{(1,1,2),(1,1,4)\}\cup\{2(1,1,c):c=3,9\}\cup\{2(1,0,c):c=2,5\}\cup\{2(2,0,3)\}.$

The statement for the nonCM case can be found in Theorem 1 of [JT]. Its proof can be found in the article of The moduli spaces of Jacobians isomorphic to a product of two elliptic curves, 2016.

Some notes: The assumption of the characteristic 0 can be removed. This was also done by Ibukiyama/Katsura/Oort.

The exact list is only known under the GRH in the nonCM case. But Kani's solution doesn't depend on the GRH.