Are the solutions to $1+1/2^s+1/3^s=0$ known?

Without even thinking I will shoot from the hip and say no explicit closed-forms are known.

Firstly, the paper you cite does not actually say the zeros have those periods. In fact, since the numbers $\log 2$ and $\log3$ are linearly independent over $\Bbb Q$, anything with both those periods would be dense on a line. Rather, zeros of $1+2^{-s}+3^{-s}$ are the intersection points of the (complex) graphs of the two functions $1+2^{-s}$ and $-3^{-s}$ (for instance), which it is easy to check have complex periods $2\pi i/\log 2$ and $2\pi i/\log 3$ respectively.

Secondly, $a^s+b^s=0$ has two exponentials but can easily be reduced to one exponential, allowing the term $s$ to be isolated. The second sum you cite is just an unravelling of another expression with two expentials, made to appear to have more exponentials artificially using the geometric sum formula. It is also noteworthy that $1+2^{-s}+3^{-s}+6^{-s}$ factors as $(1+2^{-s})(1+3^{-s})$ and so reduces to two cases each involving only one exponential.

By contrast, perturbing a two-exponential expression by a constant is not amenable to the above type of factorization, nor is it amenable to being rescaled into just one exponential. If there were known symbolic closed-form means for solving this, or if there was a special function invented for this purpose, I'd imagine it'd be more widely known. (Much like how the Lambert W function was designed to take on situations where a single-exponential equation has been perturbed by a power of the unknown under certain conditions.)