Does this 'alternating' Euler product converge for all $\Re(s) > 0$?

As Jacob mentioned, the question comes down to the conditional convergence of $\sum p^{-s}(-1)^n$ (for essentially the reason that $\prod(1-x)$ converges and is nonzero if $\sum x$ converges and $\sum x^2$ absolutely converges.

Now, if you look in Titchmarsh's Theory of Functions, there is a theorem that the region of conditional convergence of a Dirichlet series is a half plane (along with part of its boundary). This is proven by showing that convergence at some $s$ gives convergence on every wedge with vertex at $s$ and otherwise strictly to the right.

Thus since the series converges for every real $x>0$ (alternating series), it also converges in the right half plane.

Taking logarithms, we arrive at the sum

$$-\sum_{n=1}^{\infty} (-1)^n \log(1-p_n^{-s}).$$ For $Re(s)>1/k$, we can approximate the logarithm with its taylor series to $2k+1$ terms with an error of $o(p_n^{-2})$ which converges absolutely. Thus, it suffices to show that for each $s>0$, the sum

$$\sum_{n=1}^{\infty} (-1)^n p_n^{-s}$$ converges. To do this, we estimate

$$p_{2n}^{-s}-p_{2n+1}^{-s} = s^{-1}\int_{p_{2n}}^{p_{2n+1}} x^{-1-s} dx = O_s(G_{2n}\cdot p_{2n}^{-1-s})$$ where we define $G_n= p_{n+1}-p_n$.

Now note that by the PNT we have $p_n\sim n\log n$ and $\displaystyle\sum_{n=X}^{2X-1} G_n = p_{2X}-p_X \sim X\log X$.

\begin{align*} \sum_{n=1}^{\infty} (-1)^n p_n^{-s}&=\sum_{n=1}^{\infty} (p_{2n}^{-s}-p_{2n+1}^{-s})\\ &\ll_s \sum_n G_{2n}\cdot p_{2n}^{-1-s}\\ &< \sum_n G_n \cdot p_n^{-1-s}\\ &\ll \sum_{m=0}^{\infty} \sum_{n=2^m}^{2^{m+1}-1} G_n (m2^m)^{-1-s}\\ &\ll \sum_{m=0}^{\infty} (m2^m)^{-s}\\ \end{align*}

Which converges for $s>0$.