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}\begin{equation} h(z)\sqrt2=\frac{\sqrt{1+z}-\sqrt{1-z^2}}{z} =\sum_{i=0}^\infty c_{2i+1} z^{2i}+\sum_{i=1}^\infty (c_i-c_{2i}) z^{2i-1}. \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.