Riemann Zeta Function's Analytic Continuation
Let's begin by presenting a simple example of representing a function with a series. Let $f(z)=\frac{1}{1-z}$ for $z\ne 1$. Recall that $f(z)$ can be represented by the geometric series
$$f(z)=\sum_{n=0}^\infty z^n \tag 1$$
for $|z|<1$. So, although $f(z)$ exists for all $z\ne 1$, its representation as given in $(1)$ is valid only when $|z|<1$.
We can represent also represent $f(z)$ by the series
$$f(z)=-\sum_{n=1}^\infty \left(\frac{1}{z}\right)^n \tag 2$$
for $|z|>1$. So, we have two representations for the same function that are valid in distinct regions of the complex $z$-plane.
Now, suppose that we represent the function denoted $\zeta(s)$ by the series
$$\zeta(s)=\sum_{n=1}^\infty \frac{1}{n^s}$$
for $\text{Re}(s)>1$. We can easily extend the definition by writing
$$\underbrace{\sum_{n=1}^{\infty}\frac{1}{n^s}}_{\zeta(s)}=\underbrace{2\sum_{n=1}^{\infty}\frac{1}{(2n)^s}}_{2^{1-s}\zeta(s)} -\sum_{n=1}^{\infty} \frac{(-1)^n}{n^s}\tag 3$$
for $\text{Re}(s)>1$. Upon rearranging $(3)$ we find
$$\zeta(s)=(1-2^{1-s})^{-1}\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s}\tag 4$$
But notice that the series on the right-hand side of $(4)$ converges for $\text{Re}(s)>0$. So, we have just developed another representation for $\zeta(s)$ that is valid in a larger region of the complex $s$-plane.
There are other series representations of the Riemann Zeta function, such as its Laurent series,
$$\zeta(s)=\frac1{s-1}+\sum_{n=1}^\infty \frac{(-1)^n\gamma_n}{n!}(s-1)^n$$
which converges for all $s\ne 1$.
And there are integral representation of $\zeta(s)$ such as
$$\zeta(s)=\frac1{\Gamma(s)}\int_0^\infty \frac{x^{s-1}}{e^x-1}\,dx$$
which converges for $\text{Re}(s)>1$ and (see 25.5.11 this reference)
$$\zeta(s)=\frac12 +\frac{1}{s-1}-2 \int_0^\infty \frac{\sin(s\arctan(x))}{(1+x^2)^{s/2}(e^{2\pi x}+1)}\,dx$$
which is valid for all $s\ne 1$.
You are not "changing" the function. You are extending the function, i.e. making a new function with a larger domain that agrees with the original function on the old domain. More precisely, you are extending the function $\sum_{n=1}^\infty n^{-s}$ whose domain is the half plane $\Re(s)>1$ to a function with domain $\mathbb C \setminus\{1\}$ that agrees with $\sum_{n=1}^\infty n^{-s}$ for all values of $s$ with $\Re (s) >1.$ It's common practice to call both $\sum_{n=1}^\infty n^{-s}$ and its analytic continuation to $\mathbb C\setminus\{1\}$ (which is a very special type of extension) by the same name $\zeta(s)$ even though they're different functions. Remember a function actually has two parts: (1) Its domain and (2) a "rule" for producing an output from any number in the domain. Don't forget about the first part.
As for whether function is somehow "defined" for $\Re (s) \le 1$ by the sum $\sum_{n=1}^\infty n^{-s},$ I take exception on two fronts. First, what do you imagine it would be defined $as$? Infinity perhaps? This would be complicated since the sum might oscillate and not limit to infinity in a meaningful way. Maybe you just imagine "is divergent" is a definition? It might help to go back to an old textbook exercise where you're asked "what's the domain of $\sqrt{x}"$ and you dutifully answer "$x\ge 0$". Why? Because (forgetting for a moment that we know about complex numbers) we couldn't get a number from the rule $\sqrt{x}$ for $x<0.$ We're in sort of the same situation with zeta expression... we can't get a number for $\sum_{n=1}^\infty n^{-s}$ for $\Re(s) \le 1$ so this expression's (largest possible) domain is $\Re(s) > 1.$
On the other hand, perhaps you didn't mean "defined" and what you really meant was "determined". So the expression $\sum_{n=1}^\infty n^{-s}$ somehow determines uniquely that the values for $\Re(s)\le1$ are divergent, or whatever you want to call their state of being. However, this is much too strong to be true. Take our example of $\sqrt{x}$ with "domain" $x\ge0.$ Can we extend this to a function on all the reals? Of course we can! In fact we can define the function to be absolutely anything we want for $x<0.$ For instance, we could just define the extension to be zero for $x<0.$ This is a perfectly valid extension of the function whose rule is $\sqrt{x}$ and whose domain is $x\ge 0$ to a function whose domain is all the real numbers. Notice that I haven't abused notation and called the new function $\sqrt{x}$ as I mentioned above is done routinely with the zeta function.
Now, you might object "but I already know how to define a square root for all real numbers. $\sqrt{-2}=2i$, not zero." Yes, this is another valid extension. You still object "but this one's better... the other was kinda arbitrary." And I think it's better and more natural too. In fact I think it's the best and most natural definition you could ask for. It's so good I'm inclined to abuse notation and call this extension $\sqrt{x}$ and consider $\sqrt{x}$ to henceforth have domain $\mathbb R$ rather than $x>0.$
So what I've hoped to show is that extending/modifying a function is pretty much an arbitrary process, but that some extensions are nicer than others. This is exactly the same thing that's going on with the zeta function. The function whose rule is $\sum_{n=1}^\infty n^{-s}$ and whose domain is $\Re(s)>1$ has a ton of extensions to larger domains, but there is a special and natural one called the analytic continuation that extends the domain to $\mathbb C\setminus\{1\}.$ It agrees with $\sum_{n=1}^\infty n^{-s}$ for $\Re(s)>1$ and also is defined outside that domain, e.g. $\zeta(0) = -1/2$ and $\zeta(-1) = -1/12.$ And like last paragraph we like this new extended function so much that we use $\zeta(s)$ to denote it, even though anybody writing an exposition on the zeta function will first write down $\zeta(s) = \sum_{n=1}^\infty n^{-s},$ referring to a restricted version of the zeta function function with domain $\Re(s)>1.$ This is because it's a nice explicit expression that serves as a prototype for the "true" analytically continued zeta function. Just like the restricted $\sqrt{x}$ for $x\ge 0$ is a easier-to-understand prototype for the extended square root function where the domain includes negative numbers (at the conceptual cost of needing to introduce complex numbers).
So there's nothing really mysterious going on here. We can extend functions to larger domains (or even modify them on their 'natural domain') in any way we want... it's just not guaranteed to be a productive enterprise. However certain extensions are natural and it is productive. The analytic continuation of the zeta function is a good example of this.
However, explaining why the analytic continuation is a good extension that we like so much would take too much more time and this answer is already long enough.