Localising a ring twice gives ring isomorphic to localising by each subset in turn

Given a ring $$R$$, suppose we have two multiplicative subsets $$S,U \subseteq R$$ (which contain 1). Write $$US = \{us \mid u \in U, s \in S\}$$ also a multiplicative subset, containing both $$S$$ and $$U$$

Writing $$\lambda: R \to R_S$$ the localisation map, so that $$\lambda(U)$$ is also a multiplicative subset. I'm trying to find a sleek proof that $$\left( R_S \right)_{ \lambda(U)} \cong R _ {SU}$$ which uses universal properties of localisation.

I'm currently trying to show that $$\left( R_S \right)_{ \lambda(U)}$$ also has the universal property:

Given $$\phi: R \to T$$ such that $$SU$$ maps to units in $$T$$, then $$S$$ maps to units in $$T$$ so we get a unique ring homomorphism $$\phi_1$$ such that the following commutes: $$\begin{array}{cccc} R & \xrightarrow{\phi} & T \\ & \searrow & \uparrow & \phi_1 \\ & & R_S \end{array}$$ and then $$\phi_1$$ maps $$\lambda(U)$$ into units of $$T$$ (since $$\phi_1 \circ \lambda(U) = \phi(U)$$ by the diagram) yielding unique $$\phi_2$$ such that $$\begin{array}{cccc} R_S & \xrightarrow{\phi_1} & T \\ & \searrow & \uparrow & \phi_2 \\ & & \left( R_S \right)_{ \lambda(U)} \end{array}$$ commutes.

This yields that we get a commutative diagram with unique $$\phi_2$$ $$\begin{array}{cccccc} R & \xrightarrow{\phi} & T & \xleftarrow{\phi_2} & \left( R_S \right)_{ \lambda(U)}\\ & \searrow & \uparrow & \nearrow \\ & & R_S \end{array}$$

So existence follows. My problem is that this contruction depends on $$\phi_1$$. This is where I'm struggling: the universal property would need uniqueness with $$\begin{array}{cccc} R & \xrightarrow{\phi_1} & T \\ & \searrow & \uparrow & \phi_2 \\ \mu \circ \lambda& & \left( R_S \right)_{ \lambda(U)} \end{array}$$

where $$\mu: R_S \to \left( R_S \right)_{ \lambda(U)}$$ is the natural localisation map. (and $$\mu \circ \lambda$$ is the diagonal arrow. I'm sorry about the label placement - I wouldn't normally use arrays but apparently AMScd doesn't have diagonal arrows)

It feels like something along the lines of this would be sufficient, but I'm not sure: given a commutative $$\begin{array}{cccccc} &R & \xrightarrow{\phi} & T \\ \lambda &\downarrow & & \uparrow & \phi_2 \\ &R_S & \xrightarrow{\mu} & \left( R_S \right)_{ \lambda(U)} \end{array}$$ then can we simply define $$\phi_1 := \phi_2 \circ \mu$$ which yields the diagram above, which is unique by our first argument, and then uniqueness of $$\phi_2$$ follows?

Finishing off from there if this is true, we use universal property of $$R_{SU}$$ and recently shown one for $$\left( R_S \right)_{ \lambda(U)}$$ that there are unique $$\phi$$,$$\psi$$ such that:

$$\begin{array}{ccccc} && R \\ & \swarrow & \downarrow & \searrow \\ R _ {SU} & \xrightarrow{\phi} & \left( R_S \right)_{ \lambda(U)} & \xrightarrow{\psi} & R _ {SU} \end{array}$$

and so again using universal property of $$R_{SU}$$, we have uniqueness of the identity of $$R_{SU}$$ so that $$1 = \psi \circ \phi$$ and similarly $$1 = \phi \circ \psi$$ yielding the isomorphism.

(I also realise showing $$\dfrac{r}{s} / \dfrac{u}{1} \mapsto \dfrac{r}{su}$$ is an isomorphism is relatively quick, but showing well-definedness and injectivity gets quite element heavy, and I'm hoping for something different.)

Your approach is correct. You have shown that for $$(R_S)_{\lambda(U)}$$ and $$\phi:R\rightarrow T$$ satisfying $$\phi(US) \subseteq R^\times$$ you find an extension $$\phi_2$$.
Now suppose that $$\psi:(R_S)_{\lambda_U}\rightarrow T$$ is another extension. Note that the composite $$\sigma:R_S \overset{\mu}\rightarrow (R_S)_{\lambda(U)} \overset{\psi}\rightarrow T$$ satisfies $$\sigma \circ \lambda = \phi$$ by assumption, so by the universal property of $$R_S$$ you have $$\sigma = \phi_1$$. But now you are done, since by the universal property of $$(R_S)_{\lambda(U)}$$ we obtain $$\psi = \phi_2$$.