1. The operation $*$ is defined by $a * b = a + b + ab$. We need to find the value of
$$\frac{1}{2} * \left( \frac{1}{3} * \left( \cdots * \left( \frac{1}{9} * \left( \frac{1}{10} * \frac{1}{11} \right) \right) \right) \right).$$

2. First, let's explore the properties of the operation $*$. We can rewrite $a * b$ as follows:
$$a * b = a + b + ab = (a + 1)(b + 1) - 1.$$
This can be verified by expanding the right-hand side:
$$(a + 1)(b + 1) - 1 = ab + a + b + 1 - 1 = ab + a + b.$$

3. Using this property, we can generalize the operation for multiple terms. For three terms $a, b,$ and $c$:
$$a * (b * c) = a * ((b + 1)(c + 1) - 1) = a + ((b + 1)(c + 1) - 1) + a((b + 1)(c + 1) - 1).$$
Simplifying this:
$$a * (b * c) = a + (b + 1)(c + 1) - 1 + a((b + 1)(c + 1) - 1) = (a + 1)((b + 1)(c + 1)) - 1.$$

4. By induction, we can show that for $n$ terms $x_1, x_2, \ldots, x_n$:
$$x_1 * x_2 * \cdots * x_n = -1 + \prod_{i=1}^n (x_i + 1).$$
Base case: For $n = 2$, we have:
$$x_1 * x_2 = (x_1 + 1)(x_2 + 1) - 1.$$
Inductive step: Assume it holds for $n$ terms. For $n+1$ terms:
$$x_1 * x_2 * \cdots * x_n * x_{n+1} = \left( -1 + \prod_{i=1}^n (x_i + 1) \right) * x_{n+1} = \left( \prod_{i=1}^n (x_i + 1) - 1 \right) * x_{n+1}.$$
Using the property of $*$:
$$\left( \prod_{i=1}^n (x_i + 1) - 1 \right) * x_{n+1} = \left( \prod_{i=1}^n (x_i + 1) \right) * x_{n+1} = \left( \prod_{i=1}^n (x_i + 1) + 1 \right)(x_{n+1} + 1) - 1 = \prod_{i=1}^{n+1} (x_i + 1) - 1.$$

5. Applying this result to our specific problem:
$$\frac{1}{2} * \left( \frac{1}{3} * \left( \cdots * \left( \frac{1}{9} * \left( \frac{1}{10} * \frac{1}{11} \right) \right) \right) \right) = -1 + \prod_{i=2}^{11} \left( \frac{1}{i} + 1 \right).$$
Simplifying the product:
$$\prod_{i=2}^{11} \left( \frac{1}{i} + 1 \right) = \prod_{i=2}^{11} \left( \frac{1 + i}{i} \right) = \prod_{i=2}^{11} \left( \frac{i + 1}{i} \right).$$
This product telescopes:
$$\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdots \frac{12}{11} = \frac{12}{2} = 6.$$
Therefore:
$$-1 + 6 = 5.$$

The final answer is $\boxed{5}$.