The Riemann zeta function $\zeta(s)$ is a sum of reciprocals of powers of natural numbers,
$$\zeta(s) = \sum_{n \geq 1} \frac{1}{n^s}.$$
As written, this makes sense for complex numbers $s$ so long as $\text{Re } s > 1$. For these numbers, there is little more to be said.
But you've asked about an interesting number: $\zeta(1 + i)$, and $\text{Re }(1 + i) \not > 1$. What's happening there is a bit subtle, and a bit abusive in terms of notation.
It turns out there is another function (let's call it $Z(s)$) which makes sense for all complex numbers $s$ except for $s = 1$, and which exactly agrees with $\zeta(s)$ when $\text{Re } s > 1$. If you're familiar with some calculus or complex analysis, then you should also know that the function $Z(s)$ is also complex differentiable everywhere except for $s = 1$. This is a very special property that distinguishes $Z(s)$. The theory of complex analysis (in particular, the theory of "analytic continuation") gives that there can be at most one function that extends $\zeta(s)$ to a larger region, like $Z(s)$ does.
In this sense, we could realize that $Z(s)$ is uniquely determined by $\zeta(s)$. As it agrees with $\zeta(s)$ everywhere $\zeta(s)$ (initially) makes sense, it might even be reasonable to just use the name $\zeta(s)$ instead of $Z(s)$. That is, when I write $\zeta(s)$, what I'm really saying is
$$\zeta(s) = \begin{cases}
\zeta(s) & \text{if Re }s > 1 \\
Z(s) & \text{otherwise }
\end{cases}$$
It is this function that W|A computes when you ask it for $\zeta(1 + i)$.
Although what I've written is true (and important), it doesn't answer one aspect of your question
What is it even calculating?
I mentioned there exists this function $Z(s)$, or rather that it is possible to give meaningful values to $\zeta(s)$ for all $s \neq 1$. But how? Stated differently, yo're asking what is the analytic continuation of the Riemann zeta function?
The continuation is unique, but the steps to get there are not. I'll give a very short, incomplete proof that describes one way to calculate $\zeta(1+i)$.
We start by considering $\displaystyle h(s) = \sum_{n \geq 1} \frac{2}{(2n)^s}$. Performing some rearrangements,
$$\begin{align}
h(s) &= \sum_{n \geq 1} \frac{2}{(2n)^s} \\
&= \frac{1}{2^{s - 1}} \sum_{n \geq 1} \frac{1}{n^s} \\
&= \frac{1}{2^{s - 1}} \zeta(s)
\end{align}$$
Let's subtract this from the regular zeta function. On the one hand,
$$ \zeta(s) - h(s) = \zeta(s)(1 - \frac{1}{2^{s-1}}).$$
On the other hand,
$$ \begin{align}\zeta(s) - h(s) &= \sum_{n \geq 1} \left( \frac{1}{n^s} - \frac{2}{(2n)^s} \right) \\
&= \sum_{n \geq 1} \frac{(-1)^{n+1}}{n^s},
\end{align}$$
and this last series makes sense for $\text{Re } s > 0$. (If you haven't looked at alternating series before, this might not be obvious. But the idea is that the sign changes cancel out a lot of the growth, so much that it converges for a larger region).
In total, this means that
$$\zeta(s) = (1 - 2^{s - 1})^{-1} \sum_{n \geq 1} \frac{(-1)^{n+1}}{n^s},$$
and you can just "plug in" $1+i$ here. [Notice that the problem when $s = 1$ is apparent here, as you cannot divide by $0$.] In practice, it's an infinite sum, so you'll take the first very many terms to get the value of $\zeta(1+i)$ to any precision you want.
For completeness, it also turns out that
$$\pi^{-s/2} \zeta(s) \Gamma(\tfrac{s}{2}) = \pi^{(s-1)/2} \zeta(1-s) \Gamma(\tfrac{1-s}{2}),$$
which lets us transform values of $\zeta(s)$ for $\text{Re } s > 0$ into values when $\text{Re } s < 1$. The $\Gamma(z)$ function here is called the "Gamma function" (it's an integral, a sort of generalization of a factorial) and this equation is called the symmetric functional equation of the zeta function.
