= Solution
The <Möbius function> has $\mu(1)=1$, vanishes on integers divisible by a square of a prime, and equals $(-1)^r$ on a product of $r$ distinct primes. Factoring the divisor sum prime by prime gives
$$
\sum_{m\mid n}\mu(m)=\begin{cases}1,&n=1,\\0,&n>1.\end{cases}
$$
This is the <Möbius divisor-sum identity>.
For $T/2\le t\le T$ and $0<\sigma\le1$, the <Hardy-Littlewood approximation to the Riemann zeta function> at cutoff $T$ gives
$$
\zeta(s)=\sum_{d\le T}d^{-s}+O(T^{-\sigma}).
$$
Indeed $T\ge|t|/\pi$, and the omitted integral term has size at most $T^{1-\sigma}/|s-1|\ll T^{-\sigma}$. Multiply by $\sum_{m\le M}\mu(m)m^{-s}$. Its absolute value is at most
$$
\sum_{m\le M}m^{-\sigma}\le M^{1-\sigma}\sum_{m\le M}\frac1m\ll M^{1-\sigma}\log M,
$$
where the elementary inequality follows from $m^{1-\sigma}\le M^{1-\sigma}$. Reindexing the finite double sum yields coefficients $a_n=\sum_{m\mid n,\ m\le M,\ n/m\le T}\mu(m)$, with no terms for $n>MT$. For $n\le\min(M,T)$ all divisors meet both restrictions, so the <Möbius divisor-sum identity> gives $a_1=1$ and $a_n=0$ for $1<n\le\min(M,T)$. Hence the <truncated Möbius inverse identity for the Riemann zeta function> is
$$
\boxed{\zeta(\sigma+it)\sum_{m\le M}\frac{\mu(m)}{m^{\sigma+it}}=1+\sum_{\min(M,T)<n\le MT}\frac{a_n}{n^{\sigma+it}}+O\left(\frac{M\log M}{T^\sigma M^\sigma}\right).}
$$
The displayed error is uniform in $0<\sigma\le1$ and the stated height interval.
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