There is NO such example.

Note that any semisimple algebraic ${\mathbb{R}}$-group $H$ of Hermitian type has a *compact* (anisotropic) maximal torus.
Indeed, by a definition of a group of Hermitian type
(see, e.g., Deligne, Travaux de Shimura, condition (1.5.3) on page 128), $H$ is an *inner form* of a compact algebraic $\Bbb R$-group $K$, namely, $H=\,_\sigma K$, where $\sigma={\rm inn}(x)\in {\rm Aut}(K)$, $x^2=1$, $x\in K^{\rm ad}(\Bbb R)$,
$K^{\rm ad}=K/Z_K$.
Let $T_K\subset K$ be a maximal $\Bbb R$-torus such that $T_K^{\rm ad}(\Bbb R)$ contains $x$, where $T_K^{\rm ad}=T_K/Z_K$.
Then $T_K=\,_\sigma T_K\subset \,_\sigma K=H$ is a compact maximal torus of $H$.

**Theorem.** Let $G$ be a (connected) reductive ${\mathbb{R}}$-group. Write $G^{\rm der}=[G,G]$. Assume that $G^{\rm der}$ has a *compact* maximal torus $T^{\rm der}$.
If the image of $Z_G({\mathbb{R}})$ in $\pi_0(G({\mathbb{R}}))$ is nontrivial,
then there exists a nontrivial split ${\mathbb{R}}$-subtorus $T'\subset Z_G$ and
a reductive ${\mathbb{R}}$-subgroup $G''\subset G$ such that $G=T'\times_{\mathbb{R}} G''$.

*Proof.*
Write $T=Z_G\cdot T^{\rm der}$; then $T$ is a maximal torus of $G$.
We have maps
$$ Z_G({\mathbb{R}})\to \pi_0(T({\mathbb{R}}))\to \pi_0(G({\mathbb{R}})).$$
Since the image of $Z_G({\mathbb{R}})$ in $\pi_0(G({\mathbb{R}}))$ is nontrivial, we have
$ \pi_0(T({\mathbb{R}}))\neq 1$.

Write $T=T_0\times_{\mathbb{R}} T_1\times_{\mathbb{R}} T_2$,
where $T_0$ is a compact ${\mathbb{R}}$-torus, $T_1$ is a split ${\mathbb{R}}$-torus, and $T_2\simeq (R_{{\mathbb{C}}/{\mathbb{R}}}{\mathbb G}_{m,{\mathbb{C}}})^{n}$.
We have $\pi_0(T_0({\mathbb{R}}))=1$, and $\pi_0(T_2({\mathbb{R}}))=1$.
Since $\pi_0(T({\mathbb{R}}))\neq 1$, we conclude that $T_1\neq 1$.
Note that the ${\mathbb{R}}$-torus $T/Z_G$ is isogenous to $T^{\rm der}$, and hence, compact.
It follows that $T_1\subset Z_G$.

Set $T'=T_1$ and $T''=T_0\times_{\mathbb{R}} T_2$; then clearly $T=T'\times_{\mathbb{R}} T''$.
Set $G''=T''\cdot G^{\rm der}$.
Then
$$T'\cdot G''=T'\cdot T''\cdot G^{\rm der}=T\cdot G^{\rm der}=G\quad \text{and}\quad T'\cap G''=T'\cap T''=1.$$
Since $T'\subset Z_G$, it commutes with $G''$.
Thus $G=T'\times_{\mathbb{R}} G''$, as required.

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