Is there an intuitive reason for a certain operation to be associative?

A common way to build weird-looking associative operations is to start from a known one, such as multiplication, say on the real numbers or some subset of them, and then to transform it through some bijection $\alpha$, by defining $$x\ast y=\alpha^{-1}(\alpha(x)\cdot\alpha(y)).$$ Indeed this is equivalent to $\alpha(x\ast y)=\alpha(x)\cdot \alpha(y)$ (so that $\alpha$ is actually an isomorphism), and it is then easy to check associativity by noticing that \begin{align*}\alpha(x\ast (y\ast z)) & =\alpha(x)\cdot \alpha(y\ast z) = \alpha(x)\cdot(\alpha(y)\cdot \alpha(z))\\ & =(\alpha(x)\cdot \alpha(y))\cdot \alpha(z) = \alpha(x\ast y)\cdot \alpha(z)\\ & =\alpha((x\ast y)\ast z),\end{align*} which implies that $x\ast (y\ast z)=(x\ast y)\ast z$ since $\alpha$ is bijective. Other properties, such as commutativity or existence of neutral or inverses, can be done in the same way, depending on the cases.

In this case, we can see that $$\frac{1}{x\ast y}=\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}$$so that $$1+\frac{1}{x\ast y}=1+\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}=\left(1+\frac{1}{x}\right)\cdot\left(1+\frac{1}{y}\right),$$ so if you define $\alpha(x)=1+\frac{1}{x}$, you can check that it defines a bijection $(0,+\infty)\to (1,+\infty)$, and $\ast$ is just a transformation of the multiplication on $(1,+\infty)$, which explain why it is associative. In fact you can also see right away that it must also be commutative, but that it can't have a neutral element (otherwise $(1,+\infty)$ would have one).

Another way to prove associativity quite easily is when your operation is already known to be commutative. Then it is enough (and in fact necessary) to find an expression $$ (x\ast y) \ast z = \varphi(x,y,z) $$ where $\varphi$ is invariant under (circular) permutation, meaning that $\varphi(t_1,t_2,t_3) = \varphi(t_{\sigma(1)},t_{\sigma(2)},t_{\sigma(3)})$ for all (circular) permutations $\sigma \in \mathfrak S_3$.

Indeed, you then get $$ (x\ast y) \ast z = \varphi(x,y,z) = \varphi(y,z,x) = (y\ast z)\ast x = x \ast (y \ast z) $$ where the last equality holds by commutativity of $\ast$. (Taking a good look at what happen above, showing that $\varphi$ is invariant under the transposition $(1\,3)$ is actually enough.)

In you case, commutativity of $\ast$ is directly given by commutativity of product and sum of reals. And the expression you found for $(x \ast y)\ast z$ is clearly invariant by permutation of $x,y,z$ (again by commutativity of products and sums of reals). So you can conclude that $\ast$ is associative.