Convergence of prime zeta function for $\mathfrak R(s)=1$?

Checking your calculations, I see a sign error, which however is inconsequential. You have

$$I(x) = O(1/\log x) + (1+i\alpha) B(x) + (1+i\alpha)I(x),$$

which gives you

$$\frac{-i\alpha}{1+i\alpha}I(x) = B(x) + O_{\alpha}(1/\log x),$$

where you wrote the factor $\dfrac{i\alpha}{1+i\alpha}$.

As mentioned before, that doesn't affect the conclusion that you have

$$S_{\alpha}(x) = C + O(1/\log x),$$

whence the convergence of the series for $s = 1+i\alpha$ when $\alpha \neq 0$.

For completeness, let's include a different proof of convergence:

By bounds obtained by Pierre Dusart, we have

$$\log n + \log \log n - 1 < \frac{p_n}{n} < \log n + \log \log n$$

for $n \geqslant 6$. Setting $t_n = n\log n$, we then have $0 < p_n - t_n < n\log \log n$ for $n \geqslant 6$, and hence

$$\lvert p_n^{-s} - t_n^{-s}\rvert = \biggl\lvert s\int_{t_n}^{p_n} \frac{du}{u^{s+1}}\biggr\rvert \leqslant \lvert s\rvert \int_{t_n}^{p_n} \frac{du}{u^{1 + \operatorname{Re} s}} < \lvert s\rvert \frac{p_n-t_n}{t_n^{1 + \operatorname{Re} s}} < \lvert s\rvert \frac{\log \log n}{n^{\operatorname{Re} s}(\log n)^{1 + \operatorname{Re} s}}.$$

We see that for $\operatorname{Re} s \geqslant 1$

$$\sum_{n = 2}^{\infty} \biggl\lvert \frac{1}{p_n^{s}} - \frac{1}{t_n^{s}}\biggr\rvert < +\infty,$$

and hence $\sum p^{-s}$ converges if and only if $\sum t_n^{-s}$ converges.

The convergence of the latter series for $s = 1 + i\alpha,\; \alpha \in \mathbb{R}\setminus \{0\}$ is relatively easy to show.

First we note that via integration by parts

$$T(m) := \sum_{n = 2}^m \frac{1}{n^s} = \frac{1}{2}\bigl(2^{-s} + m^{-s}\bigr) + \int_2^m \frac{dt}{t^s} - s\int_2^m \frac{\{t\} -\tfrac{1}{2}}{t^{s+1}}\,dt,$$

where $\{t\}$ denotes the fractional part of $t$. The last integrand is Lebesgue-integrable on $[2,+\infty)$, and the first integral is $\frac{m^{1-s} - 2^{1-s}}{1-s}$, which, since $1-s$ is purely imaginary, is bounded by $\frac{2}{\lvert \operatorname{Im} s\rvert}$. So there is a $C\in \mathbb{R}$ with $\lvert T(m)\rvert \leqslant C\tag{1}$ for all $m$.

Next, for $0 < a < b$ we have

$$\biggl\lvert \frac{1}{a^s} - \frac{1}{b^s}\biggr\rvert = \biggl\lvert s\int_a^b \frac{dt}{t^{s+1}}\biggr\rvert \leqslant \lvert s\rvert \int_a^b \frac{dt}{\lvert t^{s+1}\rvert} = \lvert s\rvert \int_a^b \frac{dt}{t^2} = \lvert s\rvert\biggl(\frac{1}{a} - \frac{1}{b}\biggr). \tag{2}$$

Now a summation by parts shows

\begin{align} \sum_{n = k}^m \frac{1}{n^s}\cdot \frac{1}{(\log n)^s} &= \sum_{n = k}^m \bigl(T(n) - T(n-1)\bigr)\frac{1}{(\log n)^s}\\ &= \sum_{n = k}^m \frac{T(n)}{(\log n)^s} - \sum_{n = k-1}^{m-1} \frac{T(n)}{\bigl(\log (n+1)\bigr)^s}\\ &= \frac{T(m)}{\bigl(\log (m+1)\bigr)^s} - \frac{T(k-1)}{(\log k)^s} + \sum_{n = k}^m T(n)\biggl(\frac{1}{(\log n)^s} - \frac{1}{\bigl(\log (n+1)\bigr)^s}\biggr). \end{align}

The triangle inequality, $(1)$ and $(2)$ now yield

\begin{align} \Biggl\lvert \sum_{n = k}^m \frac{1}{n^s}\cdot \frac{1}{(\log n)^s}\Biggr\rvert &\leqslant \frac{C}{\log (m+1)} + \frac{C}{\log k} + C\lvert s\rvert \sum_{n = k}^m \biggl(\frac{1}{\log n} - \frac{1}{\log (n+1)}\biggr)\\ &= \frac{C(1+\lvert s\rvert)}{\log k} - \frac{C(\lvert s\rvert - 1)}{\log (m+1)}\\ &< \frac{C(1+\lvert s\rvert)}{\log k}, \end{align}

so the series

$$\sum_{n = 2}^{\infty} \frac{1}{t_n^s} = \sum_{n = 2}^{\infty} \frac{1}{n^s(\log n)^s}$$

is convergent. By the introductory remark, it follows that

$$\sum_p \frac{1}{p^s}$$

is convergent for $\operatorname{Re} s = 1$ and $s \neq 1$.