For a group $G=\mathbb{Z}_{p^r}^k$ this is quite straightforward.
Let $g$ have order $p^r$ in $G$ (if not then we are effectively working
in $\mathbb{Z}_{p^s}$ where $s < k$). Applying an automorphism of $G$
we can assume that $g=(1,0,\ldots,0)$. Endomorphisms of $G$ correspond
to matrices over $\mathbb{Z}_{p^r}$. The image of $s$ is the first row
(or column if you put the map on the other side) and we see that the image
of $s$ is uniformly distibuted over all of $G$.

Now consider a general finite abelian $p$-group $G$. There
is a surjective homomorphism $\pi:H\to G$ where $H=\mathbb{Z}_{p^r}^k$
for some $r$ and $k$. Each endomorphism of $G$ lifts to an endomorphism
of $H$ and the number of these lifted endomorphisms is the same
for each endomorphism of $G$. Thus the images of $g\in G$ under
an arbitrary endomorphism of $G$ are the images under $\pi$
of the images of a lifting $h$ of $h$ under an arbitrary endomorphism of
$H$. Morevover these images are uniformly distibuted over all
admissible elements. The upshot is that the random endomorphism of $G$
takes $g$ to $\pi(h')$ where $\pi(h)=g$ and $h'\in H$ has the same
order as $h$ and all such images are acheived with equal probability.

For general finite abelian $G$ split up $G$ as a product of its Sylow
$p$-subgroups. Then $g\in G$ splits up into its primary components
and each of these behave in the same way, under a random endomorphism,
as in the $p$-group case above.