The Harish-Chandra projection is multiplicative on the center

By Central elements lie in the zero-weight subspace of $U(\mathfrak{g})$, both $z$ and $z^{\mathrm{\prime }}$ lie in $U(\mathfrak{g}{)}_0$. Write $z=\mathrm{pr}(z)+a$ and $z^{\mathrm{\prime }}=\mathrm{pr}(z^{\mathrm{\prime }})+a^{\mathrm{\prime }}$ with $a,a^{\mathrm{\prime }}\in {\mathfrak{n}}^{-}U(\mathfrak{g})+U(\mathfrak{g}){\mathfrak{n}}^{+}$.

Put $I:={\mathfrak{n}}^{-}U(\mathfrak{g})+U(\mathfrak{g}){\mathfrak{n}}^{+}$. Multiplication on either side by $U(\mathfrak{h})$ preserves $I$. Moreover, if $c$ is central, then $Ic\subseteq I$: for a term $xu$ with $x\in {\mathfrak{n}}^{-}$, one has $(xu)c=x(uc)$, while for a term $uy$ with $y\in {\mathfrak{n}}^{+}$, centrality gives $(uy)c=(uc)y$.

Using only the decomposition of $z$, write $z{z}^{\mathrm{\prime }}=\mathrm{pr}(z)z^{\mathrm{\prime }}+a{z}^{\mathrm{\prime }}=\mathrm{pr}(z)\mathrm{pr}(z^{\mathrm{\prime }})+\mathrm{pr}(z)a^{\mathrm{\prime }}+a{z}^{\mathrm{\prime }}\mathrm{.}$ The last two terms lie in $I$ by step 2.1; no assertion that $I$ is multiplicatively closed is needed.

Applying the projection from The Harish-Chandra projection to step 3.1 leaves exactly $\mathrm{pr}(z)\mathrm{pr}(z^{\mathrm{\prime }})$. Therefore $\mathrm{pr}(z{z}^{\mathrm{\prime }})=\mathrm{pr}(z)\mathrm{pr}(z^{\mathrm{\prime }})$.