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 Zeta function of $\mathbb{F}_q[T]$ is $$\sum_{f \text{ monic}} x^{\deg f} = \sum_{n \ge 0} q^n x^n = \frac{1}{1-qx}.$$ The Euler product identity 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: Let $\mu: \mathbb{F}_q[T] \to \mathbb{C}$ be the polynomial Möbius function, defined by $$\mu(f) = \begin{cases} 0 & f \text{ not squarefree} \\ (-1)^k & f=p_1 \cdots p_k (p_i \text{ distinct irreducibles}) \end{cases}.$$ 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 $$\sum_{f \text{ monic, squarefree of degree }n} \mu(f).$$ Since $\mu(f)=0$ for $f$ which is not squarefree, 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. It should be compared with the following equivalent formulation of the Riemann Hypothesis: $$\sum_{n \le x } \mu(n) = O(x^{\frac{1}{2}+\varepsilon}),$$ where this time $\mu$ is the integer Möbius function.