I have found a solution for functional equation $F(x+1) = x + F(x) \tag1$
$F(x) = \frac{x(x-1)}{2}+C$, where $C$ - some constant
However, I have also tried to recursively expand equation $(1)$ at $0$:
$F(0) = -1 + F(-1) = -1 -2 + F(-2) = ... = -1 -2 -3 -4 -5 ... = -\sum_{n=1}^\infty {n} = -\zeta(-1)$
From that I conclude that $C = -\zeta(-1) = 1/12$
Can we say that $F(x) = \frac{x(x-1)}{2}+\frac{1}{12}$ is the only possible solution, or should we also append ", given zeta-function has analytical expansion to negative values"?
No, you cannot conclude this. Each step of the way in your $\ldots$, you carry with you a copy of some $F(-n)$, which contains the same $C$ as $F(0)$ does on the other side of the equality sign. Taking the limit (which in this case is dubious in the first place, as the limit in the conventional sense doesn't exist) does not make that $C$ disappear, and indeed, any constant $C$ will actually work. A direct inspection will immediately show that no matter what $C$ is, the functional equation is still satisfied.
More explicitly, if we swap out $F$ for your expression, we get $$ \frac{0(0-1)}{2} + C = -1 + \frac{-1(-1-1)}{2} + C\\ = -1-2 + \frac{-2(-2-1)}{2} + C\\ = -1-2-3 + \frac{-3(-3-1)}2 + C\\ \vdots\\ = -1-2-3-\cdots - n + \frac{-n(-n-1)}2 + C\\ \vdots $$ The fact that you push the ${}+C$ further and further to the right as you go on doesn't mean that it disappears in the limit (even if we were allowed to take the limit). It's still there.