I think you're approaching the question in the wrong way. The whole point is that you can show that for $\Re(s) > 1$, $$\Phi(s) = \sum_{p}{\frac{\log p}{p^s}} = \frac{1}{s - 1} + E(s),$$ where the function $E(s)$ is holomorphic meromorphic on the larger open half-plane $\Re(s) > 1/2$ ; with poles possibly at the zeroes of $\zeta(s)$; in fact, Zagier shows that $$E(s) = - \frac{\zeta'(s)}{\zeta(s)} - \frac{1}{s - 1} - \sum_{p}{\frac{\log p}{p^s (p^s - 1)}}.$$
Basically, this is saying that $\Phi(s)$ is meromorphic on an open neighbourhood of $\Re(s) > 1/2$ \geq 1$ with a simple pole at $s = 1$, and the expansion above shows that the residue of $\Phi(s)$ at $s = 1$ is equal to $1$. This is precisely the same as saying that $$\lim_{\varepsilon \to 0} \varepsilon \Phi(1 + \varepsilon) = 1.$$ Indeed, we have that $$\lim_{\varepsilon \to 0} \varepsilon \Phi(1 + \varepsilon) = \lim_{\varepsilon \to 0} \frac{\varepsilon}{1 + \varepsilon - 1} + \lim_{\varepsilon \to 0} \varepsilon E(1 + \varepsilon),$$ and the first limit tends to $1$ (obviously) while the second limit tends to zero (as $E(1 + \varepsilon)$ tends to something finite).
If you don't understand this method at all (i.e. all about meromorphic extensions of functions, poles, and residues), then this is probably due to a lack of background in complex analysis. Seeing as this proof of the prime number theorem is all about complex analysis, I'd recommend reading up on all these basics beforehand.
