The following $100$ numbers are written on the board: $$2^1 - 1, 2^2 - 1, 2^3 - 1, \dots, 2^{100} - 1.$$

Alice chooses two numbers $a,b,$ erases them and writes the number $\dfrac{ab - 1}{a+b+2}$ on the board. She keeps doing this until a single number remains on the board.

If the sum of all possible numbers she can end up with is $\dfrac{p}{q}$ where $p, q$ are coprime, then what
is the value of $\log_{2}(p+q)$?
1. **Define the function and verify symmetry:**
   Let \( f(a, b) = \frac{ab - 1}{a + b + 2} \). We need to verify that \( f(a, b) = f(b, a) \):
   \[
   f(a, b) = \frac{ab - 1}{a + b + 2} = \frac{ba - 1}{b + a + 2} = f(b, a)
   \]
   This shows that \( f \) is symmetric.

2. **Prove the associativity property:**
   We need to show that \( f(f(a, b), c) = f(f(a, c), b) = f(f(b, c), a) \). Let's compute \( f(f(a, b), c) \):
   \[
   f(a, b) = \frac{ab - 1}{a + b + 2}
   \]
   Let \( x = f(a, b) \). Then,
   \[
   f(x, c) = \frac{xc - 1}{x + c + 2}
   \]
   Substituting \( x = \frac{ab - 1}{a + b + 2} \):
   \[
   f(f(a, b), c) = \frac{\left(\frac{ab - 1}{a + b + 2}\right)c - 1}{\frac{ab - 1}{a + b + 2} + c + 2}
   \]
   Simplifying the numerator:
   \[
   \left(\frac{ab - 1}{a + b + 2}\right)c - 1 = \frac{abc - c - (a + b + 2)}{a + b + 2} = \frac{abc - c - a - b - 2}{a + b + 2}
   \]
   Simplifying the denominator:
   \[
   \frac{ab - 1}{a + b + 2} + c + 2 = \frac{ab - 1 + (a + b + 2)(c + 2)}{a + b + 2} = \frac{ab - 1 + ac + bc + 2a + 2b + 4}{a + b + 2}
   \]
   Thus,
   \[
   f(f(a, b), c) = \frac{abc - a - b - c - 2}{ab + ac + bc + 2a + 2b + 2c + 3}
   \]
   Similarly, we can show that:
   \[
   f(f(b, c), a) = \frac{abc - a - b - c - 2}{ab + ac + bc + 2a + 2b + 2c + 3}
   \]
   Therefore, \( f(f(a, b), c) = f(f(b, c), a) \), proving the associativity property.

3. **Prove the specific case:**
   We need to show that if \( a = 2^k - 1 \) and \( b = 2^{k+1} - 1 \), then \( f(a, b) = \frac{1}{2^{k+1} - 1} \):
   \[
   f(a, b) = \frac{(2^k - 1)(2^{k+1} - 1) - 1}{(2^k - 1) + (2^{k+1} - 1) + 2}
   \]
   Simplifying the numerator:
   \[
   (2^k - 1)(2^{k+1} - 1) - 1 = 2^{k+k+1} - 2^k - 2^{k+1} + 1 - 1 = 2^{2k+1} - 2^k - 2^{k+1}
   \]
   Simplifying the denominator:
   \[
   (2^k - 1) + (2^{k+1} - 1) + 2 = 2^k + 2^{k+1} - 2 + 2 = 2^k + 2^{k+1}
   \]
   Thus,
   \[
   f(a, b) = \frac{2^{2k+1} - 2^k - 2^{k+1}}{2^k + 2^{k+1}} = \frac{2^{2k+1} - 2^k - 2^{k+1}}{2^k(1 + 2)}
   \]
   Simplifying further:
   \[
   f(a, b) = \frac{2^{2k+1} - 2^k - 2^{k+1}}{2^k \cdot 3} = \frac{2^{2k+1} - 2^k - 2^{k+1}}{3 \cdot 2^k} = \frac{2^{k+1} - 1}{3}
   \]
   This shows that \( f(a, b) = \frac{1}{2^{k+1} - 1} \).

4. **Induction and final value:**
   Using induction and the associativity property, we can show that the final value after repeatedly applying \( f \) is \( \frac{1}{2^{100} - 1} \).

5. **Sum of all possible numbers:**
   Since there is only one final value, the sum of all possible numbers is \( \frac{1}{2^{100} - 1} \).

6. **Calculate \( \log_2(p + q) \):**
   Given \( \frac{p}{q} = \frac{1}{2^{100} - 1} \), we have \( p = 1 \) and \( q = 2^{100} - 1 \). Thus,
   \[
   p + q = 1 + 2^{100} - 1 = 2^{100}
   \]
   Therefore,
   \[
   \log_2(p + q) = \log_2(2^{100}) = 100
   \]

The final answer is \( \boxed{100} \)