Yes, it is true.
Your expression is the coefficient of $x^n$ in the following product: $$\prod_{P\text{ monic irreducible}} (1-x^{\deg P}) = \prod_{n} (1-x^n)^{M(n)}.$$ The Euler product identity (for the Zeta function of $\mathbb{F}_q[T]$) tells us that: $$\frac{1}{1-qx} = \prod_{P\text{ monic irreducible}} (1-x^{\deg P})^{-1} =\prod_{n} (1-x^n)^{-M(n)}.$$ Taking its reciprocal, we find that $$1-qx = \prod_{P\text{ monic irreducible}} (1-x^{\deg P}).$$ Now it is just a matter of comparing coefficients on both sides.
Interpretation: The term $\prod_{i=1}^{s} \binom{M(n_i)}{e_i}$ counts the number of monic, squarefree polynomials of degree $n$ whose factorization consists of $e_i$ irreducibles of degree $n_i$. The (polynomial) Möbius function $\mu(\bullet)$ assumes the value $(-1)^{\sum e_i}$ for each such polynomial. In other words, your sum may be rewritten as $$\sum_{f \text{ monic, squarefree of degree }n} \mu(f).$$ Note that $\mu(f)=0$ for $f$ which is not squarefree. In other words, your claim is the same as $$n>1 \implies \sum_{f \text{ monic of degree }n} \mu(f) = 0.$$ This is classical, and due to L. Carlitz, whose proof was exactly as above.