My previous answer, in much simplified and more explicit form: take $g(z):=\frac{1-\sqrt{1-z}}{z}$ and $$h(z):=f(z)g(z)=\frac{1-\sqrt{1-z}}{z}\,\sqrt{\frac{1+z}{2}}
=\frac{\sqrt{1+z}-\sqrt{1-z^2}}{z\sqrt2}.$$ 
Then $g$ and $h$ are pgf's. 

Indeed, as before, let 
\begin{equation}
	c_j:=\frac1{j2^{2j-1}} \binom{2 j-2}{j-1}>0. 
\end{equation} 
Then 
\begin{equation}
	\sqrt{1+z}=1+\sum_{j=1}^\infty(-1)^{j-1}c_j z^j,\quad 
	\sqrt{1-z^2}=1-\sum_{i=1}^\infty c_i z^{2i},
\end{equation}
\begin{equation}
	g(z)=\frac{1-\sqrt{1-z}}{z}=\sum_{j=1}^\infty c_j z^{j-1}, 
\end{equation}
\begin{equation}
	h(z)=\frac{\sqrt{1+z}-\sqrt{1-z^2}}{z\sqrt2}
	=\sum_{i=0}^\infty c_{2i+1} z^{2i+1}+\sum_{i=1}^\infty (c_i-c_{2i}) z^{2i}. 
\end{equation}

It remains to check that $c_i\ge c_{2i}$ for $i\ge1$. Let $r_i:=c_{2i}/c_i$. Then $r_{i+1}/r_i=\frac{16i^2-1}{16i^2-4}>1$ for $i\ge1$, so that $r_i$ is increasing in $i\ge1$ to $\lim_{i\to\infty}r_i=\frac1{2\sqrt2}<1$. So, $r_i<1$ for $i\ge1$, which confirms that $c_i\ge c_{2i}$. This completes the proof. 

I am retaining the previous answer, because it shows some of the process by which the second answer was obtained.