Suppose $\alpha = a/b$ is rational. All but finitely many primes are relatively prime to $b$. For these primes, we have $$ e^{ 2\pi i ap/b} = \sum_{\chi: (\mathbb Z/b)^\times \to \mathbb C^\times} \chi(p) \frac{ \sum_{x \in \mathbb Z/b^\times} \overline{\chi}(x) e^{ 2\pi i ax/b} }{ \phi(b)}$$
so $$\frac{1}{ 1- e^{ 2\pi i ap/b}p^{-s} } = \prod_{\chi: (\mathbb Z/b)^\times \to \mathbb C^\times} \left( \frac{1}{1 - \chi(p) p^{-s} }\right)^{\frac{ \sum_{x \in \mathbb Z/b^\times} \overline{\chi}(x) e^{ 2\pi i x/b} }{ \phi(b)}} + O(p^{-2s})$$
and thus your Euler product is $$ \prod_{\chi: (\mathbb Z/b)^\times \to \mathbb C^\times} L (\chi,s) ^{\frac{ \sum_{x \in \mathbb Z/b^\times} \overline{\chi}(x) e^{ 2\pi i ax/b} }{ \phi(b)}} $$
where we take $L(1,s)$ to be the Riemann zeta function, times something holomorphic on $\sigma>1/2$. In particular, it has a pole at $s=1$ of order $ (1/\phi (b))\sum_{x \in \mathbb Z/b^\times}e^{2\pi i a/b}$, which for $a$ relatively prime to $b$ is $\mu(b)/\phi(b)$ (Ramanujan sum).
In particular, if $b$ is squarefree so $\mu(b) \neq 0$, this has a pole or zero at $s=1$ of non-integer order, hence isn't holomorphic.
If $b$ is squarefree, its holomorphicity in the half-space $\sigma > 1/2$ depends on the Riemann hypothesis for these Dirichlet $L$-functions and the Riemann zeta function.