Is the “sum of all natural numbers” unique?

What if we demand the functions $f_n$ to be analytic rather than just continuous?

No problem. Define

$$f_n(s) = \frac{n}{(n - (-1)^n)^s},$$

where $k^s$ is as usual defined using the real value of $\log k$ (works since $n - (-1)^n > 0$). Then $f_n(0) = n$ for all $n$, and by a standard argument the series converges absolutely and locally uniformly for $\operatorname{Re} s > 2$. We compute \begin{align} \sum_{n = 1}^{\infty} f_n(s) &= \sum_{n = 1}^{\infty} \frac{n}{(n - (-1)^n)^s} \\ &= \sum_{n = 1}^{\infty} \Bigl(\frac{1}{(n - (-1)^n)^{s-1}} + \frac{(-1)^n}{(n - (-1)^n)^s}\Bigr) \\ &= \sum_{n = 1}^{\infty} \frac{1}{(n - (-1)^n)^{s-1}} + \sum_{n = 1}^{\infty} \frac{(-1)^n}{(n - (-1)^n)^s} \\ &= \biggl(\frac{1}{2^{s-1}} + \frac{1}{1^{s-1}} + \frac{1}{4^{s-1}} + \frac{1}{3^{s-1}} + \ldots\biggr) \\ &\qquad + \biggl(-\frac{1}{2^s} + \frac{1}{1^s} - \frac{1}{4^s} + \frac{1}{3^s} - \frac{1}{6^s} + \frac{1}{5^s} - \ldots\biggr) \\ &= \sum_{m = 1}^{\infty} \frac{1}{m^{s-1}} + \sum_{m = 1}^{\infty} \frac{(-1)^{m-1}}{m^s} \\ &= \zeta(s-1) + \eta(s) \end{align} for $\operatorname{Re} s > 2$. This has an analytic continuation to $\mathbb{C}\setminus \{2\}$, and the value at $0$ is $$\zeta(-1) + \eta(0) = -\frac{1}{12} + \frac{1}{2} = \frac{5}{12}.$$

One can by similar means obtain different values.

Such summation methods are however very ad-hoc, as far as I know every "reasonable" summation method assigns either $+\infty$ (the natural value) or $-\frac{1}{12}$ to the divergent series. I admit that I don't know a good definition of "reasonable" for summation methods (except maybe "extends 'limit of partial sums', is linear and stable", but that definition excludes several widely used summation methods).