The "universal theory" question is easy to solve: indeed if $G$ is a group and $H$ a subgroup, any universal formula true for $G$ is true for $H$. Hence if $G,H$ are groups and both embed into each other, then $G,H$ have the same universal theory.

Now let $G$ be $F_2\times F_2$ and $H$ the kernel of the homomorphism $G\to\mathbf{Z}$ mapping all four generators to $1$. Then $H$ is not finitely presented, and contains a copy of $G$. So $G,H$ have the same universal theory.

(However they are not EE since $H$ doesn't satisfy the formula expressing: there are elements $x_1,\dots,x_4$ such that each element is uniquely the product of an element in the centralizer of $\{x_1,x_2\}$ and one in the centralizer of $\{x_3,x_4\}$.)

The "universal theory" question is easy to solve: indeed if $G$ is a group and $H$ a subgroup, any universal formula true for $G$ is true for $H$. Hence if $G,H$ are groups and both embed into each other, then $G,H$ have the same universal theory.

Denote by $F_2$ the free group on two generators.

Now let $G$ be $F_2\times F_2$ and $H$ the kernel of the homomorphism $G\to\mathbf{Z}$ mapping all four generators to $1$. Then $H$ is not finitely presented, and contains a copy of $G$. So $G,H$ have the same universal theory.

(However they are not EE since $H$ doesn't satisfy the formula expressing: there are elements $x_1,\dots,x_4$ such that each element is uniquely the product of an element in the centralizer of $\{x_1,x_2\}$ and one in the centralizer of $\{x_3,x_4\}$.)