We can use induction on $n\geq 3$ to prove associativity of the system repair rate $\overline{\mu}$. The associativity of the system failure rate $\overline{\lambda}=\lambda_1+\cdots\lambda_n$ is already given, since addition is associative.
At first we note that for $n\geq 1$ the expressions
\begin{align*}
\mu_{1,2,\ldots,n}&:=\left(\sum_{j=1}^n\lambda_j\right)
\frac{\prod_{k=1}^n\frac{\mu_k}{\mu_k+\lambda_k}}{1-\prod_{l=1}^k\frac{\mu_l}{\mu_l+\lambda_l}}\\
\lambda_{1,2,\ldots,n}&:=\sum_{j=1}^n\lambda_j
\end{align*}
are both symmetric in $1,2,\ldots,n$, so that each index-permutation $\sigma(1),\sigma(2),\ldots,\sigma(n)$ yields the same result. Therefore we consider wlog
\begin{align*}
\mu_{n^{\star}}&:=\mu_{1,2,\ldots,n}\\
\lambda_{n^{\star}}&:=\lambda_{1,2,\ldots,n}
\end{align*}
with index-tuple $(1,2,\ldots,n)$ always in this order.
Base step: $n=3$
We have
\begin{align*}
\mu_{2^{\star}}&=\left(\lambda_1+\lambda_2\right)
\frac{\frac{\mu_1}{\mu_1+\lambda_1}\,\frac{\mu_2}{\mu_2+\lambda_2}}
{1-\frac{\mu_1}{\mu_1+\lambda_1}\,\frac{\mu_2}{\mu_2+\lambda_2}}\tag{2}\\
\lambda_{2^{\star}}&=\lambda_1+\lambda_2\\
\\
\mu_{3^{\star}}&=\left(\lambda_1+\lambda_2+\lambda_3\right)
\frac{\frac{\mu_1}{\mu_1+\lambda_1}\,\frac{\mu_2}{\mu_2+\lambda_2}\,\frac{\mu_3}{\mu_3+\lambda_3}}
{1-\frac{\mu_1}{\mu_1+\lambda_1}\,\frac{\mu_2}{\mu_2+\lambda_2}\,\frac{\mu_3}{\mu_3+\lambda_3}}\tag{3}\\
\lambda_{3^{\star}}&=\lambda_1+\lambda_2+\lambda_3\\
\end{align*}
In the base step it is sufficient due to symmetry to show
\begin{align*}
\mu_{2^{\star},3}=\mu_{3^{\star}}\tag{4}
\end{align*}
We obtain from (2) and (4)
\begin{align*}
\mu_{2^{\star},3}&=\left(\lambda_{2^{\star}}+\lambda_3\right)
\frac{\frac{\mu_{2^{\star}}}{\mu_{2^{\star}}+\lambda_{2^{\star}}}\,\frac{\mu_3}{\mu_3+\lambda_3}}
{1-\frac{\mu_{2^{\star}}}{\mu_{2^{\star}}+\lambda_{2^{\star}}}\,\frac{\mu_3}{\mu_3+\lambda_3}}\tag{5}
\end{align*}
Since
\begin{align*}
\color{blue}{\frac{\mu_{2^{\star}}}{\mu_{2^{\star}}+\lambda_{2^{\star}}}}
&=\frac{\left(\lambda_1+\lambda_2\right)
\frac{\frac{\mu_1}{\mu_1+\lambda_1}\,\frac{\mu_2}{\mu_2+\lambda_2}}
{1-\frac{\mu_1}{\mu_1+\lambda_1}\,\frac{\mu_2}{\mu_2+\lambda_2}}}
{\left(\lambda_1+\lambda_2\right)
\frac{\frac{\mu_1}{\mu_1+\lambda_1}\,\frac{\mu_2}{\mu_2+\lambda_2}}
{1-\frac{\mu_1}{\mu_1+\lambda_1}\,\frac{\mu_2}{\mu_2+\lambda_2}}+\left(\lambda_1+\lambda_2\right)}\\
&=\frac{
\frac{\frac{\mu_1}{\mu_1+\lambda_1}\,\frac{\mu_2}{\mu_2+\lambda_2}}
{1-\frac{\mu_1}{\mu_1+\lambda_1}\,\frac{\mu_2}{\mu_2+\lambda_2}}}
{\frac{\frac{\mu_1}{\mu_1+\lambda_1}\,\frac{\mu_2}{\mu_2+\lambda_2}}
{1-\frac{\mu_1}{\mu_1+\lambda_1}\,\frac{\mu_2}{\mu_2+\lambda_2}}+1}\\
&=\frac{\frac{\mu_1}{\mu_1+\lambda_1}\,\frac{\mu_2}{\mu_2+\lambda_2}}
{\frac{\mu_1}{\mu_1+\lambda_1}\,\frac{\mu_2}{\mu_2+\lambda_2}+\left(1-\frac{\mu_1}{\mu_1+\lambda_1}\,\frac{\mu_2}{\mu_2+\lambda_2}\right)}\\
&\,\,\color{blue}{=\frac{\mu_1}{\mu_1+\lambda_1}\,\frac{\mu_2}{\mu_2+\lambda_2}}\tag{6}
\end{align*}
we obtain by putting (6) into (5)
\begin{align*}
\color{blue}{\mu_{2^{\star},3}}
&=\left(\lambda_{2^{\star}}+\lambda_3\right)
\frac{\frac{\mu_{2^{\star}}}{\mu_{2^{\star}}+\lambda_{2^{\star}}}\,\frac{\mu_3}{\mu_3+\lambda_3}}
{1-\frac{\mu_{2^{\star}}}{\mu_{2^{\star}}+\lambda_{2^{\star}}}\,\frac{\mu_3}{\mu_3+\lambda_3}}\\
&=\left(\lambda_1+\lambda_2+\lambda_3\right)
\frac{\frac{\mu_1}{\mu_1+\lambda_1}\,\frac{\mu_2}{\mu_2+\lambda_2}\,\frac{\mu_3}{\mu_3+\lambda_3}}
{1-\frac{\mu_1}{\mu_1+\lambda_1}\,\frac{\mu_2}{\mu_2+\lambda_2}\,\frac{\mu_3}{\mu_3+\lambda_3}}\\
&\,\,\color{blue}{=\mu_{3^{\star}}}
\end{align*}
and the base step follows.
Induction hypothesis: $n=N-1$
We assume the claim is valid for $n=N-1$, i.e. we have
\begin{align*}
\mu_{(N-1)^{\star},N}&=\left(\lambda_{(N-1)^{\star}}+\lambda_N\right)
\frac{\frac{\mu_{(N-1)^{\star}}}{\mu_{(N-1)^{\star}}+\lambda_{(N-1)^{\star}}}\,\frac{\mu_N}{\mu_N+\lambda_N}}
{1-\frac{\mu_{(N-1)^{\star}}}{\mu_{(N-1)^{\star}}+\lambda_{(N-1)^{\star}}}\,\frac{\mu_N}{\mu_N+\lambda_N}}\\
&=\left(\sum_{j=1}^N\lambda_j\right)\,
\frac{\prod_{k=1}^N\frac{\mu_k}{\mu_k+\lambda_k}}{1-\prod_{l=1}^N\frac{\mu_l}{\mu_l+\lambda_l}}\tag{7}\\
&=\mu_{N^{\star}}
\end{align*}
Induction step: $n=N$
We have
\begin{align*}
\mu_{N^{\star},N+1}&=\left(\lambda_{N^{\star}}+\lambda_{N+1}\right)
\frac{\frac{\mu_{N^{\star}}}{\mu_{N^{\star}}+\lambda_{N^{\star}}}\,\frac{\mu_{N+1}}{\mu_{N+1}+\lambda_{N+1}}}
{1-\frac{\mu_{N^{\star}}}{\mu_{N^{\star}}+\lambda_{N^{\star}}}\,\frac{\mu_{N+1}}{\mu_{N+1}+\lambda_{N+1}}}\tag{8}\\
\end{align*}
Since according to the induction hypothesis
\begin{align*}
\color{blue}{\frac{\mu_{N^{\star}}}{\mu_{N^{\star}}+\lambda_{N^{\star}}}}
&=\frac{\left(\sum_{j=1}^N\lambda_j\right)\,
\frac{\prod_{k=1}^N\frac{\mu_k}{\mu_k+\lambda_k}}{1-\prod_{l=1}^N\frac{\mu_l}{\mu_l+\lambda_l}}}
{\left(\sum_{j=1}^N\lambda_j\right)\,
\frac{\prod_{k=1}^N\frac{\mu_k}{\mu_k+\lambda_k}}{1-\prod_{l=1}^N\frac{\mu_l}{\mu_l+\lambda_l}}+\left(\sum_{j=1}^N\lambda_j\right)}\\
&=\frac{\frac{\prod_{k=1}^N\frac{\mu_k}{\mu_k+\lambda_k}}{1-\prod_{l=1}^N\frac{\mu_l}{\mu_l+\lambda_l}}}
{\frac{\prod_{k=1}^N\frac{\mu_k}{\mu_k+\lambda_k}}{1-\prod_{l=1}^N\frac{\mu_l}{\mu_l+\lambda_l}}+1}\\
&=\frac{\prod_{k=1}^N\frac{\mu_k}{\mu_k+\lambda_k}}
{\prod_{k=1}^N\frac{\mu_k}{\mu_k+\lambda_k}+\left(1-\prod_{k=1}^N\frac{\mu_k}{\mu_k+\lambda_k}\right)}\\
&\,\,\color{blue}{=\prod_{k=1}^N\frac{\mu_k}{\mu_k+\lambda_k}}\tag{9}
\end{align*}
we obtain by putting (9) into (8)
\begin{align*}
\color{blue}{\mu_{N^{\star},N+1}}&=\left(\lambda_{N^{\star}}+\lambda_{N+1}\right)
\frac{\frac{\mu_{N^{\star}}}{\mu_{N^{\star}}+\lambda_{N^{\star}}}\,\frac{\mu_{N+1}}{\mu_{N+1}+\lambda_{N+1}}}
{1-\frac{\mu_{N^{\star}}}{\mu_{N^{\star}}+\lambda_{N^{\star}}}\,\frac{\mu_{N+1}}{\mu_{N+1}+\lambda_{N+1}}}\\
&=\left(\sum_{j=1}^{N+1}\lambda_j\right)
\frac{\prod_{k=1}^{N+1}\frac{\mu_k}{\mu_k+\lambda_k}}
{1-\prod_{k=1}^{N+1}\frac{\mu_k}{\mu_k+\lambda_k}}\\
&\,\,\color{blue}{=\mu_{(N+1)^{\star}}}
\end{align*}
and the claim follows.
Best Answer
You already established that $$\hat \lambda = \sum_{i=1}^m \lambda_i w_i,$$ where $$w_i = \frac{u_i}{\sum u_j}$$ is a weighting factor representing the proportion of uptime observed for machine $i$. Moreover, you also established $$\hat \tau = \frac{1}{\hat \lambda}, \quad \tau_i = \frac{1}{\lambda_i}.$$ So all that remains to do is to write $$\hat \tau = \frac{1}{\hat \lambda} = \left(\sum_{i=1}^m \frac{w_i}{\tau_i}\right)^{-1}.$$ This is the desired relationship.
The reason why there is no formula of the form $$\hat \tau = \sum_{i=1}^m \tau_i w_i'$$ for alternative suitable weights $w_i'$ is because the mean times between failures for each individual machine $\tau_i$ is a reciprocal rate, where time is in the numerator. Other examples of reciprocal rates might be:
Consequently, you cannot add two reciprocal rates together to get a meaningful quantity--for instance, if car A takes $1$ minute to travel a mile, and car B takes $2$ minutes to travel a mile, it makes no sense to say their average time to travel a mile is $1 + 2 = 3$ minutes. Instead, you have to observe that car A's rate is $1$ mile per minute, and car B's rate is $0.5$ miles per minute, and so their average rate is equal to some weighted average, where the weighting is the proportion of time each car travels.