The keywords here are associativity and commutativity. The operation $x \circ y = \frac{xy}{x + y}$ clearly has the property $x \circ y = y \circ x$. Less obviously, it also has the property
$$(x \circ y) \circ z = x \circ (y \circ z).$$
You can verify this by computing both sides to be $\frac{xyz}{xy + yz + zx}$. Once you know this, it follows that $x \circ y$ defines a commutative semigroup (on, say, the positive reals, to ensure that the division is always well-defined), and in a commutative semigroup the expression
$$x_1 \circ x_2 \circ ... \circ x_n$$
is well-defined without the need to insert parentheses (by associativity) and moreover independent of the order of the $x_i$ (by commutativity).
One slick way to see that $\circ$ is indeed associative and commutative is to write it as
$$x \circ y = \frac{1}{ \frac{1}{x} + \frac{1}{y} } = f^{-1}(f(x) + f(y))$$
where $f$ (say, as a function from the positive reals to the positive reals) denotes inversion. Then the associativity and commutativity of $\circ$ follows from the associativity and commutativity of $+$, since
$$x \circ y = f^{-1}(f(x) + f(y)) = f^{-1}(f(y) + f(x)) = y \circ x$$
and
$$\begin{align*} (x \circ y) \circ z &=& f^{-1}(f(f^{-1}(f(x) + f(y))) + f(z)) \\\
&=& f^{-1}(f(x) + f(y) + f(z)) \\\
&=& f^{-1}(f(x) + f(f^{-1}(f(y) + f(z)))) \\\
&=& x \circ (y \circ z). \end{align*}$$
It follows by induction that
$$x_1 \circ ... \circ x_n = f^{-1}(f(x_1) + ... + f(x_n)) = \frac{1}{ \frac{1}{x_1} + ... + \frac{1}{x_n} }.$$
You can think of this as saying that $\circ$ is just another name for the addition operation, but it's being encoded in a funny way by $f$ and you need to decode it by $f^{-1}$ for everything to make sense. More formally, $f$ defines an isomorphism from the positive reals under $\circ$ to the positive reals under addition.
By writing down other choices for $f$ you can write down more complicated examples of functions with the same property using transport of structure. For example, if $f(x) = \log x$, then $f^{-1}(x) = e^x$, and
$$f^{-1}(f(x) + f(y)) = e^{\log x + \log y} = e^{\log x} e^{\log y} = xy$$
so $f$ defines an isomorphism from, say, positive reals under multiplication to the reals under addition (both of which define groups).
As another example, if $f(x) = x^2$, then $f^{-1}(x) = \sqrt{x}$, and
$$f^{-1}(f(x) + f(y)) = \sqrt{x^2 + y^2}$$
so $f$ defines an isomorphism from, say, the non-negative reals under the operation $x \circ y = \sqrt{x^2 + y^2}$ to the non-negative reals under addition.
As a final example, let $f(x)$ be the inverse hyperbolic arctangent, so that $f^{-1}(x) = \tanh(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}}$. Then it turns out that
$$f^{-1}(f(x) + f(y)) = \frac{x + y}{1 + xy}$$
which is (in appropriate units) the velocity-addition formula in special relativity, and $f$ defines an isomorphism from the interval $(-1, 1)$ under the above operation to the reals under addition.
