Showing $Z(H\wr K)$, for abelian $H\neq 1$ and arbitrary $K$, is the diagonal subgroup of the base group. This is Exercise 1.6.14 of Robinson's "A Course in the Theory of Groups (Second Edition)". According to Approach0 and this search, it is new to MSE.
The Details:
The definition of the wreath product given in the book is quite intricate.
On page 32 and 33, ibid.,

Let $H$ and $K$ be permutation groups on sets $X$ and $Y$ respectively. [. . .]
If $\gamma\in H$, $y\in Y$, and $\kappa\in K$, define
$$\gamma(y):\begin{cases} (x,y) \mapsto (x\gamma, y) & \\ (x,y_1) \mapsto (x,y_1) & y_1\neq y,\end{cases}$$
and
$$\kappa^\ast: (x,y)\mapsto (x,y\kappa).$$
[. . .] The functions $\gamma\mapsto \gamma(y)$, with $y$ fixed, and $\kappa\mapsto \kappa^\ast$ are monomorphisms from $H$ and $K$ to ${\rm Sym}(X\times Y)$: let their images be $H(y)$ and $K^\ast$, respectively. Then the wreath product of $H$ and $K$ is the permutation group on $X\times Y$ [. . .]
$$H\wr K=\langle H(y), K^\ast\mid y\in Y\rangle.$$
[. . .]
[T]he base group $B$ of the wreath product [is]
$$B=\underset{y\in Y}{{\rm Dr}}\, H(y).$$

Here $\underset{y\in Y}{{\rm Dr}}\, H(y)$ is defined as in this question.

Standard Wreath Products:
On page 41 of the book,

If $H$ and $K$ are arbitrary groups, we can think of them as permutation groups on their underlying sets via the right regular representation and form their wreath product $W=H\wr K$: this is called the standard wreath product.

The Question:

Let $W=H\wr K$ be the standard wreath product of an abelian group $H\neq 1$ and an arbitrary group $K$. Prove that the centre of $W$ equals the set of elements in the base group all of whose components are equal. (This is called the diagonal subgroup of the base group.)

Thoughts:
I'm confused.
I'll work through an example.
Let $H=\Bbb Z_3$ and $K=D_3$, the dihedral group of order $6$. Then $W=H\wr K$ is given by viewing $H$ and $K$ acting on their underlying sets, like so:
$$h_H^\rho: x\mapsto xh\text{ and }k_K^\rho: x\mapsto xk.$$
Let $y\in \Bbb Z_3$ be fixed. I need to specify $\gamma(y)$ and $\kappa^\ast$, for $\gamma\in \Bbb Z_3$ and $\kappa\in D_3$, as in the definition above.
What do I do next?
The diagonal subgroup for $\Bbb Z_3\wr D_3$ is unclear to me, let alone for $H,K$ as in the question.
I chose $\Bbb Z_3$ because it is cyclic and $D_3$ because it is the smallest nonabelian group. These might be bad choices, seeing as though the standard wreath product, according to GroupNames, has order $162$.

Please help :)
 A: I’m going to use the following description of $H\wr K$: the “base group” is $H^K$, the direct product of $|K|$ copies of $H$, indexed by $K$. An element will “look like” $(h_k)_{k\in K}$. It has its usual group structure.
We let $K$ act on $H^K$ as follows: given $r\in K$, $(h_k)_{k\in K}\in H^K$, we let $(h_k)^r = (h_{kr})_{k\in K}$. That is, $K$ is acting on the indices of $H^K$.
Then $H\wr K = H^K\rtimes K$, using this action.
So the elements look like pairs $((h_k)_{k\in K}, r)$ with $h_k\in H$, $r\in K$, and product
$$\bigl( (h_k)_{k\in K},r\bigr)\bigl( (g_k)_{k\in K},s\bigr) = \bigl( (h_k)_{k\in K}((g_k)_{k\in K})^{r^{-1}}, rs\bigr) = \bigl( (h_kg_{kr})_{k\in K},rs\Bigr).$$
I’m sure it won’t be hard to translate this description into yours (it’s possible one may need to switch the action from a left action to a right action, or change the shuffling as being done by $r^{-1}$ instead of $r$; sorry, but I don’t have the mental bandwidth right now to try to reconcile my usual picture of the wreath product with the description you are giving...)
First, we prove that if $((h_k)_{k\in K},r)$ is central, then $r=e$. Indeed, let $h\neq e_H$, and consider the element $\mathbf{h}\in H^K$ which has $h$ in the $e_K$ coordinate, and $e_H$ elsewhere. Then
$$(\mathbf{h},e_K)\bigl((h_k)_{k\in K},r\bigr)$$
has second coordinate $r$, and first coordinate equal to $h_k$ in the coordinates $k\in K$, $K\neq e$, and equal to $hh_e$ in the $e$ coordinate (because $e_K$ acts trivially on $(h_k)_{k\in K}$).
On the other hand,
$$\bigl( (h_k)_{k\in K},r\bigr)(\mathbf{h},e_K)$$
has first coordinate that has $h_{r^{-1}}h$ in the $r^{-1}$th coordinate, and $h_k$ in all $k$th coordinates with $k\neq r^{-1}$ (because $r^{-1}$ acted on $\mathbf{h}$ by moving $h$ to the $r^{-1}$ coordinate, and all other coordinates are trivial). Since we assumed that $((h_k),r)$ was central, the only way these two products are equal is if $r=e$. Thus, all central elements are in the “base” group, $H^K$.
Now let $((h_k),e)$ be an element of the base. Assume that there exist $k_1\neq k_2$ in $K$ with $h_{k_1}\neq h_{k_2}$. Then the product
$$((e),k_2^{-1}k_1)((h_k),e)$$
will have a first coordinate which is a tuple with $h_{k_1}$ in the $k_2$ coordinate; whereas
$$( (h_k),e) ((e),k_2^{-1}k_1)$$
will have $h_{k_2}$ in the $k_2$-coordinate of the base element. Thus, this element is not central.
So the only possible central elements are those of the form $((h)_{k\in K},e)$, where $(h)_{k\in K}$ is the tuple that has $h$ in all coordinates. Then it is a matter of verifying that they are central. But indeed, we have
$$( (h)_{k\in K},e)\bigl( (h_k),r\bigr) = \bigl( (hh_k)_{k\in K},r\bigr)$$
and
$$\bigl( (h_k)_{k\in K},e\bigr)( (h)_{k\in K},e) = \bigl( (h_kh)_{k\in K},r\bigr)$$
(because the “constant tuple” doesn’t care about the reshuffling), and because $H$ is assumed to be abelian, these two elements are equal.
(You can probably guess from this that for an arbitrary $H$, the center will instead consist of the elements with trivial $K$ coordinate, and where the base element is a constant tuple with entry in $Z(H)$ )
A: You can cut the size of your example down to  $48$ by replacing  $\mathbb Z_3$ by $\mathbb Z_2$.  But this is irrelevant.
There are apparently at least a couple different formulations of the wreath product.   Reconciling them is a thought.
More to the point, since in both formulations I've seen the base group is $H^{|K|}$, let's go with that.   Now it's pretty clear that the the diagonal in a product of a group with itself  $n $-times is the set of $n $-tuples where all the entries are the same. That is $H^{|K|}\supset\Delta=\{\underbrace{(h,h,\dots,h)}_{\text {|K|-times}}:h\in H\}$.
Of course,  since the wreath product $H\wr K $ is actually a certain semi-direct product,  we need to adjust by considering  $\Delta×\{e\}\subset H^{|K|}\rtimes K $.
Now all that is left is to understand why this is the center.  One inclusion is clear.  For now I leave the other to you, and that brings us back to understanding the definition.
