I was playing around with the following series $$S(x) = \frac{1}{x^2}+\frac{1}{x^3}+\frac{1}{x^5}+\frac{1}{x^7}+\frac{1}{x^{11}}+...=\sum_{p\in primes}\frac{1}{x^p}$$ for $x\in\mathbb{R}$ and $|x|>1$.
And it seems, that for all $x\ge2$ the value of this sum is pretty close numerically to the following expression: $$S(x) \approx \big(\sqrt{2}-1\big)\exp\big[-2(\psi(x) + \gamma-1)\big]$$ where $\psi(x)$ is digamma function and $\gamma$ is Euler–Mascheroni constant.
Is this just a coincidence or is there some explanation for this observation? And why does it work only for $x\ge 2$?
EDIT:
I'd like to show how I got my approximation. Because even though Nilotpal Sinha's answer shows that sharper estimate of $S(x)$ doesn't look like $e^{-2\psi(x)}$, I think there's something more than just coincidence of two terms in the Laurent series as $x\to\infty$ (maybe just another coincedence?..) because numerical agreement is much better.
So, if we calculate logarithm of the ratio of $\frac{S(n)}{S(n+1)}$ for $n\ge2$ we get the following: $$\log\big(\frac{S(2)}{S(3)}\big)=0.99886...\approx1$$ $$ \log\big(\frac{S(3)}{S(4)}\big)=0.65714...\approx\frac{2}{3}$$ $$ \log\big(\frac{S(4)}{S(5)}\big)=0.493396...\approx\frac{1}{2}$$ $$ \log\big(\frac{S(4)}{S(5)}\big)=0.39565...\approx\frac{2}{5}$$ And so on...
And so, for example: $$S(3) = S(2) \frac{S(3)}{S(2)} \approx S(2)e^{-1}$$ And: $$S(5) = S(2) \frac{S(3)}{S(2)}\frac{S(4)}{S(3)}\frac{S(5)}{S(4)} \approx S(2)e^{-1-\frac{2}{3}-\frac{1}{2}-\frac{2}{5}}=S(2)e^{-2\big(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\big)}=S(2)e^{-2(H_4-1)}$$ where $H_n$ is n-th Harmonic number.
Noticing that $S(2)=0.41468...\approx\sqrt{2}-1$ and generalising $H_n$ to $\psi(x)$ we get $$S(x) \approx \big(\sqrt{2}-1\big)\exp\big[-2(\psi(x) + \gamma-1)\big]$$
Your observation is a coincidence and is a consequence of the fact that the Laurent series expansion of $e^{-2\psi(x)}$ about the point $x=\infty$ is
$$ e^{-2\psi(x)} = \frac{1}{x^2} + \frac{1}{x^3} + O\bigg(\frac{1}{x^{4}}\bigg) \tag 1 $$
and its first two terms have exponents $2$ and $3$ happen to be primes. If you replace $\psi(x)$ with any other function whose first tow terms are $2$ and $3$ you will get the same observation so there is nothing special that $\psi(x)$ is doing here.
A sharper estimate of $S(x)$ will help explain this coincidence further as shown below. Using the basic properties of primes we can get remarkably sharper estimates for $S(x)$. Every primes $\ge 5$ is of the form $6k \pm 1$. Sum up the geometric sequences $x^{-6k-1} + x^{-6k+1}$ for $k = 1,2,\ldots$ and add $x^{-2} + x^{-3}$, and taking advantage of the fact the the density of primes among the first few numbers of these form is high we get
$$ S(x) = \frac{x^7 + x^6 + x^4 + x^2 -x - 1}{x^3(x^6 - 1)} + O\bigg(\frac{1}{x^{25}}\bigg) \tag 2 $$
With this much sharper estimate you can see that $S(x)$ looks nothing like $e^{-2\psi(x)}$.