Demonstration for the equal number of odd and unequal partitions of an integer

We already know from OPs derivation \begin{align*} F(x)=F\left(x^2\right)=\cdots=F\left(x^{2^q}\right)\qquad\qquad q\in\mathbb{N}_0\tag{1} \end{align*}

We set $F(x)=\sum_{n=0}^\infty a_nx^n$ and obtain from (1) by coefficient comparison \begin{align*} F(x)&=F\left(x^2\right)\\ \sum_{n=0}^{\infty}a_nx^n&=\sum_{n=0}^{\infty}a_nx^{2n}\qquad\Longrightarrow\qquad a_{2k+1}=0\qquad\forall k\geq 0\tag{2} \end{align*}

We see the coefficients of odd powers of $x$ in $F(x)$ are zero. Let's have a look at \begin{align*} &a_0+a_1x+a_2x^2+\color{blue}{a_3x^3}+\cdots\\ &\qquad=a_0+a_1x^2+a_2x^4+\color{blue}{a_3x^6}+\cdots\\ &\qquad=a_0+a_1x^4+a_2x^8+\color{blue}{a_3x^{12}}+\cdots\\ &\qquad=a_0+a_1x^8+a_2x^{16}+\color{blue}{a_3x^{24}}+\cdots\\ &\qquad\ \ \vdots \end{align*}

We observe by iteratively using $F(x)=F\left(x^2\right)=F\left(x^4\right)=\cdots$ that since for instance $a_3=0$, also all coefficients \begin{align*} a_6&=[x^6]F(x)=[x^6]F\left(x^2\right)=[x^3]F(x)=0\\ a_{12}&=[x^{12}]F(x)=[x^{12}]F\left(x^4\right)=[x^3]F(x)=0\\ a_{24}&=[x^{24}]F(x)=[x^{24}]F\left(x^8\right)=[x^3]F(x)=0\\ &\ \ \vdots \end{align*} and in general all coefficients of the form \begin{align*} a_{3\cdot2^q}=[x^{3\cdot 2^q}]F(x)=[x^{3\cdot 2^q}]F(x^{2^q})=[x^3]F(x)=0\qquad\qquad q\geq 0 \end{align*} are zero. Here we use the coefficient of operator $[x^m]$ to denote the coefficient of $x^m$ of a series.

We conclude: Each natural number $m\geq 1$ has a unique representation \begin{align*} \color{blule}{m=2^q\cdot r}\qquad\qquad q\geq 0, r\ \text{odd} \end{align*} as product of a power of $2$ (if $q\geq 1$) with an odd natural number $r$. It follows from (1) and (2) \begin{align*} \color{blue}{[x^{m}]F\left(x^{2^q}\right)}=[x^{2^{q}r}]F\left(x^{2^q}\right)=[x^r]F(x)\color{blue}{=0}\qquad\qquad q\geq 0 \end{align*}

Since from OPs representation of \begin{align*} F(x)=\prod_{k=1}^{\infty}\left(\left(1-x^{2k-1}\right)\left(1+x^k\right)\right) \end{align*} we also know that $[x^0]F(x)=1$, the claim \begin{align*} \color{blue}{F(x)=1} \end{align*} follows.