Strictly speaking, is it true that $\zeta(-1)\ne1+2+3+\cdots$?

You only have

$$\zeta(s)=\sum_{n=1}^\infty n^{-s}$$

for $\mathfrak R(s)>1$. The right-hand side of the equation is not defined otherwise.

Like you said, $\zeta(s)$ is defined by analytic continuation on the rest of the complex numbers, so the formula $\zeta(s)=\sum_{n=1}^\infty n^{-s}$ is not valid on $\mathbb C \setminus \{z\in \mathbb C, \mathfrak R(z)>1\}$.

Therefore,

$$\frac{-1}{12}=\zeta(-1)\ne \sum_{n=1}^\infty n \quad\text{(which $=+\infty$ in the best case scenario)}.$$

So what you say is correct.