They are equivalent.

If any object of $\textrm{add}(X)$ satisfies (1) then so does $X$, and $A$ satisfies (1), so (2) implies (1).

If $G$ satisfies (1) then let $I$ be the set of homomorphisms $\alpha:G\to A$, let $G^{(I)}$ be the direct sum of copies of $G$ indexed by $I$, and let $\beta:G^{(I)}\to A$ be the map which restricts to $\alpha$ on the copy of $G$ corresponding to $\alpha$. Then $\beta$ is surjective, or (1) is contradiced by the natural map $A\to A/\textrm{im}(\beta)$ and the zero map.

Taking a splitting map $\gamma:A\to G^{(I)}$ of $\beta$, the image of $\gamma$ is contained in $G^{(J)}$ for some finite $J\subseteq I$, and so $A$ is a direct summand of $G^{(J)}$ and is therefore in $\textrm{add}(G)$. So (1) implies (2).

They are equivalent.

If any object of $\textrm{add}(X)$ satisfies (1) then so does $X$, and $A$ satisfies (1), so (2) implies (1).

If $G$ satisfies (1) then let $I$ be the set of homomorphisms $\alpha:G\to A$, let $G^{(I)}$ be the direct sum of copies of $G$ indexed by $I$, and let $\beta:G^{(I)}\to A$ be the map which restricts to $\alpha$ on the copy of $G$ corresponding to $\alpha$. Then $\beta$ is surjective, or (1) is contradicted by the natural map $A\to A/\textrm{im}(\beta)$ and the zero map.

Taking a splitting map $\gamma:A\to G^{(I)}$ of $\beta$, the image of $\gamma$ is contained in $G^{(J)}$ for some finite $J\subseteq I$, and so $A$ is a direct summand of $G^{(J)}$ and is therefore in $\textrm{add}(G)$. So (1) implies (2).