Is there a real number $r$ such that $\sum\limits_{k=0}^{\infty}\frac{p_k}{r^k}=e$? Let $p_n$ denote the sequence of prime numbers, with $p_0=2$.
I'm looking for a real number $r$ such that $\sum\limits_{k=0}^{\infty}\frac{p_k}{r^k}=e$.
It's easy to show that $r>5$, with $\frac{2}{5^0}+\frac{3}{5^1}+\frac{5}{5^2}=2.8>e$.
I suppose it shouldn't be too hard to show that $r<6$.
So we know that $5<r<6$ (my observation shows that $r\approx5.7747052$).
But does it prove that there must be some value of $r\in\mathbb{R}$ such that $\sum\limits_{k=0}^{\infty}\frac{p_k}{r^k}=e$?

Write $f(x) = \sum_k p_k x^k$. Assume first that the series for $f(1/4)$ converges. Then the function $f(x)$ is defined and continuous on $[0,1/4)$. You've shown $f(1/5) > e$, and clearly $f(0) = 2$. By the intermediate value theorem, there must be some $c \in (0,1/5)$ such that $f(c) = e$. You can take $r = 1/c$.
It remains to show that $f(1/4)$ is finite. The convergence of this series will follow from the root test if we can establish, for instance, that $p_n < 3^n$. I imagine that there is an easier way to show this, but at minimum this follows by induction from Bertrand's postulate.
I don't believe the general arguments you gave are sufficient unless the series is shown to converge at some point $d$ for which $f(d) \geq e$, which will inevitably involve some estimate of the growth of $p_n$. 
