First a comment:
In most scenarios, it is more helpful to view the periodic zeta function as the polylogarithm, $\mathrm{Li}_s(z)$, evaluated at $z=e^{2i\pi x}$. The functional equation you have above comes from the functional equation for the polylogarithm on this Wikipedia page:
$$\mathrm{Li}_s(z) = \frac{\Gamma(1 - s)}{(2\pi)^{1-s}}\left(i^{1-s}~\zeta\!\left(1-s,~\frac12+{\frac{\ln(-z)}{2\pi i}}\right)+i^{s-1}~\zeta\!\left(1-s,~\frac12-{\frac{\ln(-z)}{2\pi i}}\right)\right) .$$
Solution to your problem:
Recall that when $n\in\mathbb N$ we can relate the Hurwitz zeta function to the Bernoulli polynomials by $$\zeta(1-n,x)=-\frac{B_n(x)}{n}.$$ Then for $s=n\in\mathbb N$ the part in parentheses in your question becomes $$\left(i^{1-s}\zeta(1-s,\lambda)+i^{s-1}\zeta(1-s,1-\lambda)\right)$$
$$=i^{1-s}\left(\zeta(1-n,\lambda)+(-1)^{n-1}\zeta(1-n,1-\lambda)\right)=-\frac{i^{s-1}}{n}\left(B_n(\lambda)+(-1)^{n-1}B_n(1-\lambda)\right).$$
But this is always zero since the Bernoulli polynomials have the following symmetry $$B_n(1-x)=(-1)^n B_n(x).$$ The fact that this factor is zero cancels the pole coming from $\Gamma(1-s)$ when $s$ is an integer.
Hope that helps,