Recurrence relations in the generating function of binary partitions Let $b(n)$ denote the number of binary partitions of $n$, that is, the number of partitions of $n$ as the sum of powers of $2$. Define
\begin{equation*}
F(x) = \sum_{n=0}^\infty b(n)x^n
= \prod_{n=0}^\infty \frac{1}{1-x^{2^{n}}}.
\end{equation*}
Clearly, $F(x)$ satisfies the functional equation
\begin{equation*}
(1-x)F(x) = F(x^2).
\end{equation*}
Then, we have recurrence relation
$b(2n+1)=b(2n)$ and $b(2n) = b(2n-2) + b(n), n\ge1$.
How to derive these recurrence relation from the given informations?
What I tried:
We have
\begin{equation*}
\sum_{n=0}^\infty b(2n)x^n
= \prod_{n=0}^\infty \frac{1}{1-x^{2^{2n}}}
\end{equation*}
and
\begin{equation*}
\sum_{n=0}^\infty b(2n+1)x^n
= \prod_{n=0}^\infty \frac{1}{1-x^{2^{2n+1}}}.
\end{equation*}
But, I didn't get those recurrence relation.

The sequence $\,b(n)\,$ is OEIS sequence A018819 "Binary partition function: number of partitions of n into powers of 2.". The FORMULA section
states:

$a(2m+1) = a(2m), a(2m) = a(2m-1) + a(m)$. Proof: If $n$ is odd there is a part of size $1$; removing it gives a partition of $n - 1$. If $n$ is even either there is a part of size $1$, whose removal gives a partition of $n - 1$, or else all parts have even sizes and dividing each part by $2$ gives a partition of $n/2$.

You asked

Then, we have recurrence relation
$b(2n+1)=b(2n)$ and $b(2n) = b(2n-2) + b(n), n\ge1$.


How to derive these recurrence relation from the given informations?

but you did not explicitly state that you wanted to use the ordinary
generating function $\,F(x).\,$ The first recurrence relation for
$\,b(2n+1)\,$ is already given in the FORMULA entry. For $\,b(2n)\,$
we also get from the entry $\,b(2n) = b(2n-1)+b(n)\,$ but we already
know that if $\,n\ge 1\,$ then $\,b(2n-1)=b(2n-2)\,$ and thus
$\,b(2n)=b(2n-2)+b(n)\,$ as the second recurrence relation.
Your attempt using the g.f. $\,F(x)\,$ was flawed, but can be fixed up.
$$ F(x) = \frac1{1-x}\prod_{n=1}^\infty \frac{1}{1-x^{2^{n}}} =
 \sum_{n=0}^\infty b(n)x^n. \tag{1}$$
Substitute $\,-x\,$ for $\,x\,$ to get
$$ F(-x) = \frac1{1+x}\prod_{n=1}^\infty \frac{1}{1-x^{2^{n}}} =
 \sum_{n=0}^\infty b(n)(-x)^n. \tag{2}$$
Notice that
$$ \frac12\Big(\frac1{1-x} + \frac1{1+x}\Big) = \frac1{1-x^2}, \quad
 \frac12\Big(\frac1{1-x} - \frac1{1+x}\Big) = \frac{x}{1-x^2}. \tag{3} $$
Add equation $(2)$ to $(1)$ and use equation $(3)$ to get the even part of $\,F(x)\,$
$$ \frac12(F(x)+F(-x)) = 
\frac1{1-x^2}\prod_{n=1}^\infty \frac{1}{1-x^{2^{n}}} =
 \sum_{n=0}^\infty \,b(2n)\,x^{2n}.\tag{4}$$
Subtract equation $(2)$ from $(1)$ and use equation $(3)$ to get the odd part
$$ \frac12(F(x)-F(-x)) = 
\frac{x}{1-x^2}\prod_{n=1}^\infty \frac{1}{1-x^{2^{n}}} =
 \sum_{n=0}^\infty \,b(2n+1)\,x^{2n+1}.\tag{5}$$
Divide both sides of equation $(5)$ by $\,x\,$ to get
$$ \frac1{1-x^2}\prod_{n=1}^\infty \frac{1}{1-x^{2^{n}}} =
 \sum_{n=0}^\infty \,b(2n+1)\,x^{2n}.\tag{6}$$
Equate coefficients in equations $(4)$ and $(6)$ to get $\,b(2n+1)=b(2n).\,$
From equation $(1)$ we get
$$ (1-x)F(x) = \prod_{n=1}^\infty \frac{1}{1-x^{2^{n}}} = F(x^2) \tag{7}$$
and use equation $(1)$ again with $\,x\,$ replaced with $\,x^2\,$ to get
$$ F(x^2) = \sum_{n=0}^\infty\, b(n)\,x^{2n} \tag{8}$$
and also
$$ (1-x)F(x) = 1+\sum_{n=1}^\infty \,(b(n)-b(n-1))\, x^n. \tag{9} $$
Using equations $(7)$,$(8)$, and $(9)$ together to get
$$ \sum_{k=0}^\infty\, b(k)\,x^{2k} =
  1+\sum_{n=1}^\infty \,(b(n)-b(n-1))\, x^n. \tag{10} $$
Equating coefficients of $\,x^n\,$ when $\,n\,$ is
even and odd gives the recurrence relations in the OEIS entry.
By the way, in equation $(4)$ replace $\,x^2\,$ with $\,x\,$ to get
$$ \frac1{1-x}F(x) = \sum_{n=0}^\infty \,b(2n)\,x^n \tag{11}$$
and since $\,b(2n+1)=b(2n)\,$ the same left side is the generating
function of $\,b(2n+1).\,$
