Yes, this is true for any cocommutative Hopf algebra.

Let me rewrite your statement using my (or actually my professor's) brand of Sweedler notation: I write $x_{(1)}\otimes x_{(2)}$ for $\Delta\left(x\right)$, without sum sign. And I denote the antipode of the Hopf algebra $H$ by $S$. Also I will denote your action by $\rightharpoonup $ rather than by $*$ (so I will write $X\rightharpoonup Y$ for your $X*Y$) because I prefer to use $*$ for convolution. So we have $X\rightharpoonup Y = X_{(1)} Y S\left(X_{(2)}\right)$ for all $X,Y\in H$.

Now, any $X,Y\in H$ satisfy

$\Delta\left(X\rightharpoonup Y\right) = \Delta\left(X_{(1)} Y S\left(X_{(2)}\right)\right) = \Delta\left(X_{(1)}\right) \Delta\left(Y\right) \Delta\left(S\left(X_{(2)}\right)\right)$ (since $\Delta$ is an algebra homomorphism),

whereas

$X\rightharpoonup \Delta\left(Y\right) = X\rightharpoonup \left(Y_{(1)} \otimes Y_{(2)}\right) = \left(X_{(1)} \rightharpoonup Y_{(1)}\right) \otimes \left(X_{(2)} \rightharpoonup Y_{(2)}\right)$

$= \left(X_{(1)}\right)_{(1)} Y_{(1)} S\left(\left(X_{(1)}\right)_{(2)}\right) \otimes \left(X_{(2)}\right)_{(1)} Y_{(2)} S\left(\left(X_{(2)}\right)_{(2)}\right) $

$= X_{(1)} Y_{(1)} S\left(X_{(2)}\right) \otimes X_{(3)} Y_{(2)} S\left(X_{(4)}\right) $

$= X_{(1)} Y_{(1)} S\left(X_{(4)}\right) \otimes X_{(2)} Y_{(2)} S\left(X_{(3)}\right) $

(since $X_{(1)} \otimes X_{(2)} \otimes X_{(3)} \otimes X_{(4)} = X_{(1)} \otimes X_{(4)} \otimes X_{(2)} \otimes X_{(3)}$, which is because $H$ is cocommutative)

$= \underbrace{\displaystyle \left(X_{(1)} \otimes X_{(2)}\right)}_{\displaystyle =\Delta\left(X_{(1)}\right)} \underbrace{\displaystyle \left(Y_{(1)} \otimes Y_{(2)}\right)}_{\displaystyle = \Delta\left(Y\right)} \underbrace{\displaystyle \left(S\left(X_{(4)}\right) \otimes S\left(X_{(3)}\right)\right)}_{\displaystyle = \Delta\left(S_{(3)}\right)\text{ (since }S\text{ is a anti-coalgebra homomorphism)}}$

$= \Delta\left(X_{(1)}\right) \Delta\left(Y\right) \Delta\left(S\left(X_{(2)}\right)\right) = \Delta\left(X\rightharpoonup Y\right)$,

qed.

This can be cast in diagram-chasing form, but that will make it totally unreadable.

What I used are the facts that the antipode of a Hopf algebra is an anti-coalgebra homomorphism, and that if a coalgebra is cocommutative, then tensors of the form $X_{(1)} \otimes X_{(2)} \otimes ... \otimes X_{(n)}$ are symmetric. Should I prove any of these?