I think I have started the following small set of proofs, but I could use a little help really proving them. $$ \text{Let } h(x)=(-1)^x \left(x^{1/x}-1\right).\text{CMRB}=\sum _{n=1}^{\infty } h(n).$$
$$\text{MKB}={\lim_{N \to \infty} }\int_1^{2 N} e^{i \pi t} t^{1/t} \, dt.$$
$$\text{Prove } \text{Re(MKB)}=\Im\left( \int_0^{\infty } \frac{h (1-i t)-h (1+i t)}{e^{2 \pi t}+1} \, dt\right)=\Im\left(\int_0^{\infty } \frac{h(1+i t)+h(1-i t)}{e^{2 \pi t}-1} \, dt\right).$$
$$\text{Let NoMKB}=\Im\left( \int_0^{\infty } \frac{h (1-i t)-h (1+i t)}{e^{2 \pi t}-1} \, dt\right).$$
$$\text{Prove NoMKB}+\text{Re(MKB)}=\text{CMRB}.$$ $$\text{Let} g(x)=x^{1/x}.$$ $$\text{Then prove Im(MKB)= }$$ $$-i{ \int _0^{\infty}\frac{g (1+i t)+g (1-i t)}{2 e^{\pi t}}}dt-\frac{i}{\pi }. $$
I know the proof that when $$g (x)=x^{1/x};\text{MKB}=-i \int_0^{\infty } \frac{g (1+i t)}{e^{\text{$\pi $t}}} \, dt-\frac{i}{\pi }.$$ It is here.
Also, how can we relate NoMKB to Im(MKB)to obtain MRB solely from MKB and vice-versa?
To see them worked out in Mathematica, go here. Similar integral proofs are cataloged here.
I got these from trial and error starting with the Abel-Plana formula,
from Wikipedia,
where, I think, $f(x)=(-1)^x(1-(1+x)^{1/(1+x)}),$ giving us
$$\text{ CMRB= }\sum _{n=0}^{\infty } f(n).$$ $$\text{MKB}={\lim_{N \to \infty} }\int_0^{2 N} f(x) dx.$$
Here is an outline of a "proof," but I will reward the bounty to a person who gives an analytic one.
$$f(x)=(-1)^x \left(1-(x+1)^{\frac{1}{x+1}}\right)$$
$$h(x)=(-1)^x \left(x^{1/x}-1\right)$$
$$g(x)=x^{1/x}$$
$$\text{CMRB}=\sum _{n=0}^{\infty } f n=\sum _{n=1}^{\infty } f n=\sum _{n=1}^{\infty } h n$$
$$(MKB+\text{some n i}/\pi)=\underset{N\to \infty }{\text{lim}}\int_1^{2 N+1} e^{\pi \text{it}} (g(t)) \, dt=\frac{i}{\pi }-i \int_0^{\infty } \frac{g (1+i t)}{e^{\pi t}} \, dt, n\in{N}.$$
$$\underset{N\to \infty }{\text{lim}}\int_1^{2 N+1} e^{\pi \text{it}} g(t) \, dt+i \int_0^{\infty } \frac{f (i t)-f (-i t)}{e^{2 \pi t}-1} \, dt=\text{CMRB}$$
$$NoMKB+\Im{(MKB+\text{some n i}/\pi)}=i\int_0^{\infty } \frac{f (i t)-f (-i t)}{e^{2 \pi t}-1} \, dt=\int_0^{\infty } \frac{i (h(1+i t)-h(1-i t))}{e^{2 \pi t}-1} \, dt, n\in{N}.$$
You can see it worked out here.
I'm not sure how you motivated yourself to this or how you progressed through it, but it feels very convoluted. Instead of working with your 3 functions $f,g,h$ let's just focus on $$f(x)=h(x+1)=e^{i\pi x} \left(1-(x+1)^{\frac{1}{x+1}}\right)$$ (I picked the principal branch for $(-1)^x$) and in which case we obviously have by your definition $${\rm CMRB}=\sum_{n=0}^\infty f(n) \, .$$
Now you go on and define $${\rm MKB}=\int_1^{\infty} e^{i \pi t} t^{1/t} \, {\rm d}t$$ which is already a problem as it does not converge. You can make it converge by going over to $t^{1/t}-1$ instead of $t^{1/t}$. But then after substituting $t=x+1$ this is nothing else than $${\rm MKB} = \int_0^\infty f(x) \, {\rm d}x \, .$$
I think this form is much more suitable as you apply Abel-Plana. If you insist on using your limiting form for the integer $2N$ you can easily see they are connected
$$\lim_{N\rightarrow \infty} \int_1^{2N} e^{i\pi t} t^{1/t} \, {\rm d}t = \lim_{N\rightarrow \infty} \int_1^{2N} e^{i\pi t} (t^{1/t}-1) \, {\rm d}t + \lim_{N\rightarrow \infty} \int_1^{2N} e^{i\pi t} \, {\rm d}t = \int_0^\infty f(x) \, {\rm d}x - \frac{2i}{\pi} \, . $$
As for your "to prove" objective $${\rm NoMKB}+\Re ({\rm MKB})={\rm CMRB}$$ this follows immediately upon taking the real part of the Abel-Plana formula $$\sum_{n=0}^\infty f(n) = \frac{f(0)}{2} + \int_0^\infty f(x) \, {\rm d}x + i \int_0^\infty \frac{f(it)-f(-it)}{e^{2\pi t}-1} \, {\rm d}t$$ since the LHS is real, $f(0)=0$ and for any complex number $z$ we have $\Re(iz)=-\Im(z)$.
Similarly you can take the imaginary part and use $\Im(iz)=\Re(z)$ so that $$\Im({\rm MKB})=\Re\left(\int_0^\infty \frac{f(-it) - f(it)}{e^{2\pi t}-1} \, {\rm d}t \right) \, .$$
Note that ${\rm NoMKB}$ is the imaginary part instead.
Regarding your other identity involving $\Re({\rm MKB})$ I think it should read $$\Re({\rm MKB}) = -\Re\left( \int_0^\infty \frac{f(it)+f(-it)}{e^{2\pi t}-1} \, {\rm d}t \right)$$ instead.