I claim that the function $G$ satisfies a few functional equations, but in order to formulate our functional equation, we need to extend $G$ to a much larger domain. One often has to extend a function to a larger domain in order for the functional equation to make sense; this is certainly the case for the functional equations involving the Gamma function since one has to extend the factorial function from $\mathbb{N}$ to $\mathbb{C}$ for such a functional equation to be sensible.
Let $H:\{z\in\mathbb{C}\mid\text{Re}(z)<0\}\rightarrow\mathbb{C}$ be the function defined by $H(z)=\sum_{k=1}^{\infty}e^{\sqrt{k}z}$. Then $H(z)=G(e^z)-1$.
Now, let $F$ be the collection of all square free positive integers. Then
$$H(z)=\sum_{k\in F}\sum_{n=1}^{\infty}e^{\sqrt{k}nz}=\sum_{k\in F}\frac{e^{\sqrt{k}z}}{1-e^{\sqrt{k}z}}=\sum_{k\in F}\frac{-1}{2}\cdot e^{\sqrt{k}z/2}\cdot\text{sech}(\sqrt{k}z/2).$$
Generalized analytic continuation of derivative
Observe that $$H'(z)=\sum_{k\in F}\frac{\sqrt{k}e^{\sqrt{k}z}}{(1-e^{\sqrt{k}z})^2}=\sum_{k\in F}\frac{\sqrt{k}}{4}\cdot\text{sech}(\frac{\sqrt{k}z}{2})^2.$$ This formula converges on $\{a+bi\mid a,b\in\mathbb{R},a\neq 0\}$, so the derivative $H'$ admits a 'generalized analytic continuation' to the complex plane except for the imaginary axis. We have $H'(z)=H'(-z)$ whenever $z$ is a complex number that is not on the imaginary axis. However, I do not know of any rigorous generalized notion of analytic continuation in which we can extend $H'$ from the half plane so that it satisfies $H'(z)=H'(-z)$, and I cannot find out
Continuation to compactification
Now, let $\iota:\mathbb{R}\rightarrow(S^1)^F$ be the function defined by letting
$\iota(\theta)=(e^{i\sqrt{k}\theta})_{k\in F}$. Then $\iota[\mathbb{R}]$ is dense in $(S^1)^F$.
Define a function $K:(-\infty,0)\times(S^1)^F\rightarrow\mathbb{C}$ by letting
$K(t,(b_k)_{k\in K})=\sum_{k\in F}\frac{b_k \exp(\sqrt{k}t)}{1-b_k\exp(\sqrt{k}t)}.$
Then the function $K$ is an extension of the function $H$ since
$H(a+bi)=K(a,(e^{i\sqrt{k}b})_{k\in F})$. Then the function $K$ is a continuous function, and $K$ is the only continuous function where $H(a+bi)=K(a,(e^{i\sqrt{k}b})_{k\in F})$ whenever $a<0$. If $(t,(b_k)_{k\in K})\in(-\infty,0)\times S_1^F$, then
define a chart $j_{(t,(b_k)_{k\in K})}:\{z\in\mathbb{C}\mid\text{Re}(z)<0\}\rightarrow (-\infty,0)\times S_1^F$ by letting
$j_{(t,(b_k)_{k\in K})}(a+bi)=(a+t,(b_k\cdot e^{i\sqrt{k}b})_{k\in K}).$ Then the composition $K\circ j_{(t,(b_k)_{k\in F})}$ is always holomorphic. Therefore, $K$ is 'holomorphic' in the sense that the composition of $K$ with each of these coordinate charts is holomorphic and the images of these coordinate charts partition the space $(-\infty,0)\times S_1^F$ into a whole bunch of half planes.
Observe that $(-\infty,0)\times S_1^F$ is a locally compact abelian semigroup. We shall write $+$ for this semigroup operation.
Now, observe that $$\frac{nz^{n}}{z^{n}-1}=\sum_{k=0}^{n-1}\frac{z}{z-\exp(\frac{2\pi i k}{n})}.$$
Theorem: $$\sum_{j=0}^{n-1}K(\mathbf{z}+(e^{2\pi ij/n})_{k\in F})=n\cdot K(n\mathbf{z})$$ whenever $\mathbf{z}\in (-\infty,0)\times S_1^F.$
Proof: We first observe that $$-K(t,(e^{i\theta}b_k)_{k\in F})=\sum_{k\in F}\frac{b_k\exp(\sqrt{k}t)}{b_k\exp(\sqrt{k}t)-e^{i\theta}}.$$
Therefore,
$$\sum_{j=0}^{n-1}-K(t,(b_k\exp(-2\pi i j/n))_{k\in F})=\sum_{k\in F}\sum_{j=0}^{n-1}\frac{b_k\exp(\sqrt{k}t)}{b_k\exp(\sqrt{k}t)-\exp(2\pi ij/n)}$$
$$=\sum_{k\in F}\frac{n(b_k\exp(\sqrt{k}t))^n}{(b_k\exp(\sqrt{k}t))^n-1}
=n\sum_{k\in F}\frac{b_k^n\exp(n\sqrt{k}t)}{b_k^n\exp(n\sqrt{k}t)-1}
=-n\cdot K(nt,(b_k^n)_{k\in F}).$$ Q.E.D.
As a consequence, $$\sum_{j=0}^{n-1}K'(\mathbf{z}+(e^{2\pi ij/n})_{k\in F})=n^2\cdot K'(n\mathbf{z})$$ whenever $\mathbf{z}\in (-\infty,0)\times S_1^F.$
