Not in general.
For instance, if $A$ is simple then this is not possible. In fact, any isogeny $$f \colon E_1 \times \cdots \times E_n \longrightarrow A$$ would give a dual isogeny $$f^{\ast} \colon A^{\ast} \longrightarrow E_1^{\ast} \times \cdots \times E_n^{\ast}$$ and the pullback in $A^{\ast}$ of any elliptic curve in $E_1^{\ast} \times \cdots \times E_n^{\ast}$ would produce elliptic curves in $A^{\ast}$. This is a contradiction: indeed $A^{\ast}$ is simple, since by assumption $A$ is simple.
Poincaré Reducibility Theorem says that any abelian variety $A$ is isogenous to a product $A_1^{n_1} \times \cdots \times A_k^{n_k}$, where the $A_j$ are simple abelian variety, pairwise non-isogenous. The integers $n_j$ are uniquely determined, whereas the factors $A_j$ are determined up to isogeny.