The Maclaurin series of $1-(1-\frac{x^2}{2} + \frac{x^4}{24})^{2/3}$ has all coefficients positive It was shown in a previous post that the Maclaurin series of $1 - \cos^{2/3} x$ has positive coefficients. There @Dr. Wolfgang Hintze: has noticed that the truncation $1- \frac{x^2}{2} + \frac{x^4}{24}$ can be substituted for $\cos x$ (  seems to be true for all the truncations). The proof is escaping me.
Thank you for your attention!
$\bf{Added:}$ Thomas Laffey in this paper directs to a proof of the fact that if $a_1$, $\ldots$, $a_n\ge 0$ then $\alpha = \frac{1}{n}$ makes the following series positive:
$$1- (\prod_{i=1}^n (1- a_i x))^{\alpha}$$
Numerical testing suggests that $\alpha = \frac{\sum a_i^2}{(\sum a_i)^2} \ge \frac{1}{n}$ works as well ( see the case $n=2$ tested here). So in our case, instead of $\alpha = \frac{1}{2}$ we can take $\alpha = \frac{2}{3}$. Clearly, this would then be the optimal value. This would be a test case for $n=2$. The result for $\cos x$ used the special properties of the function ( solution of a certain differential equation of second order). Maybe $1- x/2 + x^2/24$ is as general as any quadratic with two positive (distinct) roots.

Hint
Consider
$$f(a)=1-\left(1-\frac{x^2}{2}+\frac{x^4}{24}\right)^a$$ and use the binomial expansion.
It should write
$$f(a)=\frac a 2 \sum_{n=1}^\infty\frac {P_n(a)}{b_n} x^{2n}$$ You should see it very quickly.
Edit
After your edit and remarks, considering
$$f(k)=1-\big[ (1-ax)(1-b x)\big]^k$$  let $b=a c$ to obtain
$$f(k)=k\sum_{n=1}^\infty \frac{a^n}{n!}\, Q_n(c)\, x^n$$ with
$$Q_1(c)=c+1 \qquad\text{and} \qquad Q_2(c)=(c^2+1)-k (c+1)^2$$ If, as you did, we choose to cancel $Q_2(c)$ that is to say to use $k=\frac{c^2+1}{(c+1)^2}$ we have
$$\color{blue}{Q_n(c)=  (n-2)\, 2^{\left\lfloor \frac{n-1}{2}\right\rfloor }\,  \frac{c\,  (c^2-1)^2}{(c+1)^{n}} \,\,R_n(c)}$$ with
$$\left(
\begin{array}{cc}
n & R_n(c) \\
 3 & 1 \\
 4 & c^2+3 c+1 \\
 5 & \left(c^2+c+1\right) \left(c^2+4 c+1\right) \\
 6 & \left(c^2+1\right) \left(c^2+4 c+1\right) \left(3 c^2+8 c+3\right) \\
 7 & \left(c^2+3 c+1\right) \left(c^2+4 c+1\right) \left(6 c^4+7 c^3+6 c^2+7   c+6\right)
\end{array}
\right)$$
