Is the coequalizer of two abelian groups in Ab the same as their coequalizer in Set?

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Let $F: \bf{Ab} \to \bf{Set}$ be the forgetful functor. Does F preserve coequalizers for all pairs of morphisms? (not just F-split pairs)

Let me restate my question. Let $f,g :G_1 \to G_2$ be any abelian group homomorphisms. Consider the equivalence relation $\sim$ on $G_2$ generated by $f(x)\sim g(x)$ for all $x \in G_1$. Does the following always hold

$x \sim y \iff x -y \in Im(f-g)$,

or do we need the splitting condition (Barr-Beck)? If this is true I suppose that it's also true for the forgetful functors on $\bf{Grp}$ and $\bf{Vect}$. I know that it's not true in general for the forgetful functor on the category of torsion-free abelian groups, as the coequalizer in $\bf{TFAb}$ imposes more relations on the quotient. Could we characterise those categories for which this is true? (if it is true at all).

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Consider the (somewhat generic) case of an element $a \in G$ of an abelian group $G$, seen as a homomorphism $\mathbb{Z} \to G$, $z \mapsto za$, paired with the trivial homomorphism $0 : \mathbb{Z} \to G$, $z \mapsto 0$. The equivalence relation generated by $za \sim 0$ can be described as follows: We have $g \sim g'$ iff $g=g'$ or $g,g' \in \langle a \rangle$. This is clearly much stronger than $g-g' \in \langle a \rangle$.

Apart from reflexive coequalizers, there are "almost no" coequalizers preserved by $\mathbf{Ab} \to \mathbf{Set}$.