I bet the product $$ \prod_{n=2}^\infty\frac 1 {1n^{s}}, $$ which is convergent for ${\rm Re}(s)>1$, has been studied before. Can it be analytically extended across the line ${\rm Re}(s)=1$? If so, is there a functional equation?
Yes, the inverse of it, $\prod_{n=2}^{\infty} \frac{n^s1}{n^s}$
has been studied in connection with the socalled $k$almost prime zeta functions of the second kind.
It can be evaluated to
$$
\prod_{n=2}^{\infty} \frac{n^s1}{n^s}=\prod_{n=1}^{\infty} \frac{(n+1)^s1}{(n+1)^s}=
\frac{1}{\prod_{j=1}^{s1}\Gamma (2e^{2\pi ij/s})}.
$$
Then one can use the properties for the $\Gamma$function. Edit: I apologize, but $s$ is assumed to be an integer above. But the question has been discussed already, see Values where infinite products of primes and composites are equal. 


Yes, It can be analytically extended across the line Re$(s)=1$. Taking the logarithm of the function and using the series expansion of $\log(1+x)$ gives you the sum $$ \sum_{k=1}^\infty \frac{\zeta(ks)1} k $$ which has a meromorphic continuation up to the line Re$(s)=0$, with poles at $s=1/k$. There is some speculation on the other page on the problem Values where infinite products of primes and composites are equal whether the continuation should go beyond the line Re$(s)=0$ but I agree with Greg Martin's comment that this is not clear. However you asked about the line Re$(s)=1$ and that follows from the above. Note though that the logarithm of the function has meromorphic continuation and this means that your function does not have meromorphic continuation since the singularity at Re$(s)=1$ will not be a pole. I doubt that the function has a functional equation. 

