No, if the center $Z(G)$ is not connected, we cannot construct $G'$ with connected $Z(G')$.

Indeed, let $\pi\colon G'\to G$ be an epimorphism of connected reductive $F$-groups with central kernel. Choose a maximal torus $T\subset G$. We have
$$
Z(G)=\ker[T\to {\rm Aut}\ {\rm Lie}(G)]=\ker[T\to {\rm Aut}\ {\rm Lie}( G_{der})].
$$
Set $T'=\pi^{-1}(T)\subset G'$, then
$$
Z(G')=\ker[T'\to {\rm Aut}\ {\rm Lie}( G'_{der})].
$$
It follows easily that $\pi(Z(G'))=Z(G)$.

Now, if $Z(G')$ is connected, then $Z(G)=\pi(Z(G'))$ must be connected.
Therefore, if $Z(G)$ is not connected, then we cannot construct $G'$ with connected center $Z(G')$.