Why the reals with the operation $x \bullet y = \sqrt[3]{x^3 + y^3}$ is a group? The operation above is a group for the real numbers, since 0 is the identity element, and the negative of any real number is its inverse, as it can be observed trivially. Associativity is less trivial, but it holds.
In fact, if we substitute 3 for any odd number (5, 7 ...), the operation satisfies the properties of the group. However, any even number fails.
Is there any geometric / analytic / ... interpretation why an operation like $x \bullet y = \sqrt[3]{x^3 + y^3}$ is associative and, as a consequence, it gives the structure of a group to the reals?

Short answer: because $\sqrt{x^2}\ne x$ for $x<0$.
Long answer, in which I prefer $\cdot$ to $\bullet$:
An operation satisfying $(x\cdot y)^n=x^n+y^n$ closes the reals, since if $n$ is odd we can take the $n$th root, & if $n$ is even we only try to take the $n$th root of something $\ge0$. And since$$((x\cdot y)\cdot z)^n=(x\cdot y)^n+z^n=x^n+y^n+z^n=(x\cdot(y\cdot z))^n,$$the operation associates. (Cancelling the power of $n$ is trivial since, even if $n$ is even, $\cdot$ is always defined to take the non-negative $n$th root anyway.) So at a minimum, we form a semigroup.
Since $x\cdot0=(x^n)^{1/n}$, for odd $n$ we also have $0$ as an identity, but for even $n$ we don't because $x\cdot0=|x|$, so it's not even a monoid, let alone a group. The last group axioms is inverses, which work for odd $n$ as you noted, but for even $n$ we have $x\cdot y\ge|x|$, so we don't have inverses either.
