It is standard in the proof of existence of inverse limits to provide a concrete construction of the inverse limit as a subring of the direct product of the system's rings satisfying the appropriate transition conditions. That is, if $\varphi_{\mu\lambda}:R^\mu\to R^\lambda$ are the transition morphisms of the system,
$\varprojlim R^\mu\cong\{(r^\mu):\underbrace{\forall\mu,\lambda,~\varphi_{\mu\lambda}(r^\mu)=r^\lambda}_{\text{transition conditions}}\}\subseteq \prod_\mu R^\mu. \tag{$\square$}$
We use $\subseteq$ to denote subring above. It is straightforward to check that the middle set is a subring; for convenience we will refer to it by $L$ (for "limit").
For good measure, you may want to convince yourself that (a) the assignment of groups of units to rings is functorial in the latter, and (b) this functor distributes through arbitrary direct products.
If $(r^\mu)\in L$ is a unit, then there is a $(s^\mu)\in L$ for which $1_{L}=(1_{R^\mu})=(r^\mu)(s^\mu)=(r^\mu s^\mu)$ and hence the coordinate $r^\mu$ is invertible for each $\mu$. Thus if $(r^\mu)\in L^\times$ then $(r^\mu)\in\prod_\mu (R^\mu)^\times$; but since $(r^\mu)$ is in $L$ it satisfies the transition conditions, and hence it is contained in the subgroup of $\prod_\mu(R^\mu)^\times$ which is isomorphic to $\varprojlim(R^\mu)^\times$ as per $(\square)$ (although working in a different category, the idea works just the same). Conversely, if $(r^\mu)$ is in the designated subgroup of $\prod_\mu(R^\mu)^\times$ (itself a subset of $\prod_\mu R^\mu$) which is isomorphic to $\varprojlim (R^\mu)^\times$, then each coordinate $r^\mu$ is a unit in $R^\mu$ with inverse say $s^\mu$, and so $(r^\mu)(s^\mu)=1_L$ shows $(r^\mu)$ is invertible and in $L$ (again since it satisfies the TCs).
Thus the copy of $\varprojlim (R^\mu)^\times$ sitting inside $\prod_\mu(R^\mu)^\times$ sitting inside $\prod_\mu R^\mu$ is equal to the units of the copy of $\varprojlim R^\mu$ sitting inside $\prod_\mu R^\mu$.