I think that the answer is yes, at least if you are working over an algebraically closed field of char $0$. Let me try to give an argument, I hope that there are no mistakes.

You have two flat families $E_1 \to C_1$ and $E_2 \to C_2$ of elliptic curves over some unknown bases $C_1, C_2$.

The first assumption says that the map $E_1 \to X$ induces a bijection on closed points. If the field is of characteristic 0, one can use Zariski's main theorem to conclude that $E_1 \to X$ is an isomorphism of varieties (see e.g. Bijection implies isomorphism for algebraic varieties). The same for $E_2 \to X$. Using these identifications $X \cong E_1$ and $X \cong E_2$, we get two flat projections $X \to C_1$ and $X \to C_2$, where the fibers are elliptic curves. In particular $C_1$ and $C_2$ must be smooth (by flat descent of smoothness).

I am interpreting the assumption that the intersection numbers are always 1 to mean that the fibers of the induced morphism $S \to C_1 \times C_2$ are singleton points, i.e. any two fibers of the families $E_1$ and $E_2$ respectively intersect in a single point (is this the precise condition you have in mind?). Since $C_1 \times C_2$ is smooth, this implies again that $X \to C_1 \times C_2$ is an isomorphism. But now by choosing a closed point $x_1 \in C_1$, we see that $C_2 = x_1 \times C_2$ is isomorphic to the fiber of $E_1 \to C_1$ at $x_1$. This means that $C_1$ is an elliptic curve. The same holds for $C_2$. Therefore $X \cong C_1 \times C_2$ is a product of elliptic curves.

I think that the answer is yes, at least if you are working over an algebraically closed field of char $0$. Let me try to give an argument, I hope that there are no mistakes.

You have two flat families $E_1 \to C_1$ and $E_2 \to C_2$ of elliptic curves over some unknown bases $C_1, C_2$.

The first assumption says that the map $E_1 \to X$ induces a bijection on closed points. If the field is of characteristic 0, one can use Zariski's main theorem to conclude that $E_1 \to X$ is an isomorphism of varieties (see e.g. Bijection implies isomorphism for algebraic varieties). The same holds for $E_2 \to X$. Using these identifications $X \cong E_1$ and $X \cong E_2$, we get two flat projections $X \to C_1$ and $X \to C_2$, where the fibers are elliptic curves. In particular $C_1$ and $C_2$ must be smooth (by flat descent of smoothness).

I am interpreting the assumption that the intersection numbers are always 1 to mean that the fibers of the induced morphism $X \to C_1 \times C_2$ are singleton points, i.e. any two fibers of the families $E_1$ and $E_2$ respectively intersect in a single point (is this the precise condition you have in mind?). Since $C_1 \times C_2$ is smooth, this implies again that $X \to C_1 \times C_2$ is an isomorphism. But now by choosing a closed point $x_1 \in C_1$, we see that $C_2 = x_1 \times C_2$ is isomorphic to the fiber of $E_1 \to C_1$ at $x_1$. This means that $C_1$ is an elliptic curve. The same holds for $C_2$. Therefore $X \cong C_1 \times C_2$ is a product of elliptic curves.

I think that the answer is yes, at least if you are working over an algebraically closed field of char $0$. Let me try to give an argument, I hope that there are no mistakes.

You have two flat families $E_1 \to C_1$ and $E_2 \to C_2$ of elliptic curves over some unknown bases $C_1, C_2$.

The first assumption says that the map $E_1 \to S$ induces a bijection on closed points. If the field is of characteristic 0, one can use Zariski's main theorem to conclude that $E_1 \to S$ is an isomorphism of varieties (see e.g. Bijection implies isomorphism for algebraic varieties). The same for $E_2 \to S$. Using these identifications, we get two flat projections $S \to C_1$ and $S \to C_2$, where the fibers are elliptic curves. In particular $C_1$ and $C_2$ must be smooth (by flat descent of smoothness).

I am interpreting the assumption that the intersection numbers are always 1 to mean that the fibers of the induced morphism $S \to C_1 \times C_2$ are singleton points, i.e. any two fibers of the families $E_1$ and $E_2$ respectively intersect in a single point (is this the precise condition you have in mind?). Since $C_1 \times C_2$ is smooth, this implies again that $S \to C_1 \times C_2$ is an isomorphism. But now by choosing a closed point $x_1 \in C_1$, we see that $C_2 = x_1 \times C_2$ is isomorphic to the fiber of $E_1 \to C_1$ at $x_1$. This means that $C_1$ is an elliptic curve. The same holds for $C_2$. Therefore $S \cong C_1 \times C_2$ is a product of elliptic curves.

I think that the answer is yes, at least if you are working over an algebraically closed field of char $0$. Let me try to give an argument, I hope that there are no mistakes.

You have two flat families $E_1 \to C_1$ and $E_2 \to C_2$ of elliptic curves over some unknown bases $C_1, C_2$.

The first assumption says that the map $E_1 \to X$ induces a bijection on closed points. If the field is of characteristic 0, one can use Zariski's main theorem to conclude that $E_1 \to X$ is an isomorphism of varieties (see e.g. Bijection implies isomorphism for algebraic varieties). The same for $E_2 \to X$. Using these identifications $X \cong E_1$ and $X \cong E_2$, we get two flat projections $X \to C_1$ and $X \to C_2$, where the fibers are elliptic curves. In particular $C_1$ and $C_2$ must be smooth (by flat descent of smoothness).

I am interpreting the assumption that the intersection numbers are always 1 to mean that the fibers of the induced morphism $S \to C_1 \times C_2$ are singleton points, i.e. any two fibers of the families $E_1$ and $E_2$ respectively intersect in a single point (is this the precise condition you have in mind?). Since $C_1 \times C_2$ is smooth, this implies again that $X \to C_1 \times C_2$ is an isomorphism. But now by choosing a closed point $x_1 \in C_1$, we see that $C_2 = x_1 \times C_2$ is isomorphic to the fiber of $E_1 \to C_1$ at $x_1$. This means that $C_1$ is an elliptic curve. The same holds for $C_2$. Therefore $X \cong C_1 \times C_2$ is a product of elliptic curves.