Let me provide a suggestion based on Pietro Majer's expansion.

If it makes it any easier, it suffices to prove the following finite version of my claim: $$\prod_{j=1}^n\frac{x}{1-a^{2j-1}}\leq\sum_{k=0}^{2n}\prod_{j=1}^k\frac{ax}{1-a^{2j}},$$ and perhaps proceed by induction on $n\geq1$.

Let me provide a suggestion based on Pietro Majer's expansion.

If it makes it any easier, it suffices to prove the following finite version of my claim: $$\prod_{j=1}^n\frac{x}{1-a^{2j-1}}\leq\sum_{k=0}^{2n}\prod_{j=1}^k\frac{ax}{1-a^{2j}},$$ and perhaps proceed by induction on $n\geq1$.

I'll give this a jumpt start for the base case $n=1$ to prove $$\frac{a^2x^2}{(1-a^2)(1-a^4)}+\frac{ax}{1-a^2}+1-\frac{x}{1-a} =\frac{a^2x^2-(1-a^4)x+(1-a^2)(1-a^4)}{(1-a^2)(1-a^4)}\geq0.$$ It suffices to verify $f(x;b)=bx^2-(1-b^2)x+(1-b)(1-b^2)\geq0$ where $0<b:=a^2<1$.

From the discriminant, $f(x;b)$ has two real roots provided $D=(1-b)(1-b^2)(1-3b)>0$; that means $0<b<\frac13$.

On the other hand, $f(x;b)$ has a minimum at $x_*=\frac{1-b^2}{2b}$. Since we anticipate $0<x_*<1$, it must be that $b^2+2b-1>0$ or $b>\sqrt{2}-1$.

The inequalities $b<\frac13$ and $b>\sqrt{2}-1$ are not compatible. hence, $f(x;b)\geq0$ for $0<a,x<1$.

Let me provide a suggestion based on Pietro Majer's expansion.

If it makes it any easier, it suffices to prove the following finite version of my claim: $$\prod_{j=1}^n\frac{x}{1-a^{2j-1}}\leq\sum_{k=0}^{2n}\prod_{j=1}^k\frac{x}{1-a^{2j}},$$ and perhaps proceed by induction on $n\geq1$.

Let me provide a suggestion based on Pietro Majer's expansion.

If it makes it any easier, it suffices to prove the following finite version of my claim: $$\prod_{j=1}^n\frac{x}{1-a^{2j-1}}\leq\sum_{k=0}^{2n}\prod_{j=1}^k\frac{ax}{1-a^{2j}},$$ and perhaps proceed by induction on $n\geq1$.