(I will use additive notation. In particular, $n\mid x$ for $n\in\mathbb N$ and $x\in G$ means there exists $y$ such that $ny=x$.)

It follows from [Szmielew’s quantifier elimination](https://mathoverflow.net/a/402100) that two elements $x,y$ of an abelian group $G$ *satisfy the same first-order formulas* if and only if
$$m\mid nx\iff m\mid ny$$
for all integers $n,m\ge0$; in other words, $x$ and $y$ have the same order in $G/mG$. This reduces to the following two special cases:
1. $x$ and $y$ have the same order (i.e., $nx=0\iff ny=0$ for all  $n$).
2. $p^l\mid p^kx\iff p^l\mid p^ky$ for all primes $p$ and $l>k\ge0$ (where we may assume that $p^k$ divides the order of $x$ and/or $y$).

Now, if $G$ is *finite*, this is equivalent to the existence of an automorphism $\phi$ such that $\phi(x)=y$, because [elementary equivalent finite structures are isomorphic](https://math.stackexchange.com/a/1518666). Moreover, in a finite (or: bounded exponent) abelian group, condition 2 implies condition 1.

In fact, this shows that your condition
$$\def\p#1{\langle#1\rangle}G/\p x\simeq G/\p y$$
is already necessary and sufficient: if it holds, then for any $m\ge0$,
$$G/(\p x+mG)\simeq(G/\p x)/m(G/\p x)\simeq(G/\p y)/m(G/\p y)\simeq G/(\p y+mG),$$
thus $|\p x+mG|=|\p y+mG|$, i.e., $x$ and $y$ have the same order in $G/mG$.

(I’m sure there is a direct algebraic proof of this using the classification of finite abelian groups and the like, but well, I’m a logician, not an algebraist.)