Given the structure of these sequences, it seems convenient to describe them as integer valued functions of binary strings rather than of natural numbers. This way, the recursions from $b$ and $c$ appear to be recursive ways of computing a quantity of a string $x$, starting from the left, resp. from the right.
Specifically, for a binary string $x:=x_r\dots x_1x_0$ of finite length $r>0$ and $m=\sum_{i=0}^rx_i2^i$, we write $f(x)$ instead of $f(m)$ for $f\in\{a,b,c\}.$ Then the recursion translate into: $$ a(x00)=a(x01)=a(x0) ,\\ a(x10)=a(x)+1,\\ a(x11)=a(x)=a(0x),\\ a(0)=0; $$
$$ b(1x)=\big(1+a(1x)\big)b(x),\\ b(0)=1. $$
In the recursion for $c$, it is convenient to write $$2^{2m+1}(2k+1)+\frac23(4^{m+1}-1)=$$$$=2^{2m+2}(k+1)+\frac23(4^m-1),$$ and $$2^{2m+1}k+\frac23(4^{m+1}-1)=$$$$= 2^{2m+1}(k+1)+\frac23(4^m-1),$$ so that there is nothing to be carried in the binary representation of the sum. Thus
$$ c((10)^n)=(n+1)!,\\ c(x1(10)^n)=(n+1)c(x(10)^n),\\ c(x00(10)^n)=c(x0(10)^n),\\ c(0)=1. $$ Finally, we may add to these $f(0x)=f(x)$ for $f\in\{a,b,c\},$ since the value of $f(x)$ only depends on the numerical value of the string.
Note that $a(x11)=a(x)$ and $a(x01)=a(x0)$ together imply $a(x1^{2p})=a(x)$ and $a(x01^{2p+1})=a(x0)$, so that in the argument of $a$, every initial block of $1$’s on the right can be removed, and in particular $a(x1)=a(x)$.
The analogous relation $b(x1)=b(x)$ then follows by induction, since it is true for $x=0$. Also, $c(x1)=c(x)$ for $c$ is a particular case of the second equation for $c$, with $n=0$. Therefore for the sake of notation we may and do prove the equality $b(x)=c(x)$ for binary strings of odd numbers.
Every such string $x$ (after dropping initial $0$’s from the left), can be written uniquely as $$x=1^{p_n}0^{q_n}\dots 1^{p_1}0^{q_1}1^{p_0},$$ with $n\ge1$ and positive integer exponents $q_i,$ $p_i$.
In this notation, $a(x)=n$, the number of blocks of $0$’s. It then immediately follows by induction that
$$ b(1^{p_n}0^{q_n}\dots 1^{p_1}0^{q_1}1^{p_0})=\prod_{i=1}^n(1+i)^{p_i}.$$
Using the relations for $c$, we can reduce the computation of $c(x)$ to the known value $c((10)^n)=(n+1)!$, extracting in order from the right to the left $p_i-1$ of the $0$’s from the block $0^{p_i}$, with no effect on the value of $c$, and $q_i-1$ of the $1$’s from the block $1^{q_i}$, with a factor $(1+i)^{p_i}$. In conclusion $$ c(1^{p_n}0^{q_n}\dots 1^{p_1}0^{q_1}1^{p_0})=\prod_{i=1}^n(1+i)^{p_i},$$ as we wished to prove.