I'd like to come back to this problem after I learn more about tetration. For now, I'm going to answer mainly as a bookmark for myself so that I can return to this problem in the future. So, this is definitely not a real answer
First, let's take a look at $\sum_{n=0}^\infty \frac{1}{2^{2^n}}$
To start, consider the sum $\sum_{n=0}^\infty x^{2^n}$. Doing a power series expansion, we can (almost) write it as
$$\sum_{n=0}^\infty x^{2^n}= \sum_{n=0}^\infty \sum_{k=0}^\infty \frac{\ln(x)^k2^{nk}}{k!} = \sum_{k=0}^\infty \frac{\ln(x)^k}{k!} \sum_{n=0}^\infty 2^{nk} = \sum_{k=0}^\infty \frac{\ln(x)^k}{k!} \frac{1}{1-2^k}$$
But, a pole shows up at $k=0$ and we've done some divergent series manipulations, which are generally invalid. However, this isn't actually a problem, how to deal with these steps can be seen here.
Anyway, this means that the actual sum is represented by
$$\int_C \frac{\csc(\pi n)}{2i} \frac{\ln(\frac{1}{x})^n}{n!} \frac{1}{1-2^n} dn$$
Where the contour $C$ is a line with $-1<\mathfrak{Re}(z) < 0$.
First, let's deal with the poles of the real line of $\frac{1}{1-2^n}$. They are represented by the series
$$\frac{i \pi}{\ln(2)} \sum_{n \in \mathbb{Z}, n \neq 0} \frac{\text{csch}\left( \frac{2 \pi^2}{\ln(2)}\right) \ln\left(\frac{1}{x}\right)^{\frac{2 \pi i}{\ln(2)}}}{\Gamma\left(1+ \frac{2 \pi i }{\ln(2)}\right)}$$
I call this part of the sum the "error", since on the real line it is purely an oscillatory series and is very small (the largest term in the series is $\approx 10^{-7}$). It's also the error between this series and a much better-behaved series and closely related series (i.e. removing this "error" term gives a nice series).
Thus, I call the next two terms the "center" (since they have the purely oscillatory part removed, so they are in some sense the center of mass of the function). The poles on the real line besides the one at $0$ contribute
$$\sum_{k=1}^\infty \frac{\ln(x)^k}{k!} \frac{1}{1-2^k}$$
And then the pole at zero contributes the curious term
$$-\frac{2 \gamma - \ln(2) + 2 \ln(\ln(\frac{1}{x}))}{2 \ln(2)}$$
Where $\gamma$ is the Euler-Masceroni constant.
Notice that when $x = \frac{1}{e}$ then the first center sum is close to
$$\sum_{k=1}^\infty \frac{(-1)^k}{k!} \frac{1}{2^k} = \frac{1-\sqrt{e}}{\sqrt{e}}$$
Thus, the "center" of $\sum_{n=1}^\infty \frac{1}{2^{2^n}}$ is equal to
$$\frac{\sqrt{2}-1}{\sqrt{2}} -\frac{2 \gamma - \ln(2) + 2 \ln(\ln(2))}{2 \ln(2)}+\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}\ln\left(2\right)^{n}}{n!}\left(\frac{1}{1-2^{n}}+\frac{1}{2^{n}}\right)$$
There's also a potentially interesting behaviour of the sum $\sum_{k=1}^\infty \frac{1}{1-2^k}$. This series is equal to
$$\sum_{k=1}^{\infty}\left(2^{-2^{-k}}-1\right)$$
Which resembles your second sum. So perhaps this could provide a connection between $S_1$ and $S_2$.
Why do I care about the center? Because, Observe that $\sum_{n=0}^\infty x^{2^n}$ has natural boundary at $|x|=1$, since it has a dense set of poles. But what about the center?
Notice that the sum
$$f(x) = -\frac{2 \gamma - \ln(2) + 2 \ln(\ln(\frac{1}{x}))}{2 \ln(2)} + \sum_{k=1}^\infty \frac{\ln(x)^k}{k!} \frac{1}{1-2^k}$$
Also has a natural boundary BUT, it has a natural extension beyond that natural boundary-- just evaluate the function for $|x|>1$. The fact that this series converges both inside and outside its natural boundary suggests to me that perhaps there is some way to write it as the limit of rational functions. Maybe I'll come back to this later.
One thing we can do, is to fix $x$ and study
$$g(q) = \sum_{k=1}^\infty \frac{\ln(x)^k}{k!} \frac{1}{1-q^k}$$
Now, our series is a Lambert series, which are deeply number-theoretic objects. Our function will then correspond to looking at a Lambert series outside its natural boundary (which, to me, is an exciting proposition). We have
$$g(q) = \sum_{m=1}^\infty q^{-m} \sum_{k|m} \frac{\ln(x)^k}{k!} $$
Letting $x = e$, then one could study the series $\sum_{k | m} \frac{1}{k!}$.
