Step 1: Recall what bijection forces.
For $f:(1,2] \to B$ to be a bijection, $B$ must equal the exact range of $f$, no more and no less. So our whole job is to find the range of $f(x) = \log_{10}(x-1)$.
Step 2: Track the input $x$.
The domain is $1 < x \le 2$. This is the interval we will push through the function step by step.
Step 3: Shift by subtracting 1.
Subtract 1 throughout the inequality $1 < x \le 2$ to get \[ 0 < x - 1 \le 1. \] So the quantity inside the logarithm lives in $(0, 1]$.
Step 4: Apply the logarithm.
Since $\log_{10}$ is increasing, it keeps the order. As $x-1$ approaches $0$ from the right, $\log_{10}(x-1) \to -\infty$, and at $x-1 = 1$ we get $\log_{10}(1) = 0$.
Step 5: Write the range.
Putting that together, \[ -\infty < f(x) \le 0, \] so the range is $(-\infty, 0]$. The right endpoint $0$ is included because $x = 2$ is in the domain.
Step 6: State $B$.
For a bijection, $B$ is exactly this range, matching option (D).
\[ \boxed{B = (-\infty,\ 0]} \]