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 recursions 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 equations 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 (i.e. for $i$ from $0$ to $n$) all $0$’s but one from  the block $0^{p_i}$, with no effect on the value of $c$, and all $1$’s but one from  the block $1^{q_i}$, multiplying by 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.