Why is an equation necessarily dimensionally correct?

It appears with the tip of @anon, I have discovered the answer to my own question.

The justification is simple; namely, linear substitutions work for integrals. Here's the idea: take $$\int_{-\infty}^\infty e^{-ax^2} dx = f(a)$$ and make the substitutions $x \to kx$ and $a \to \frac{a}{k^2}$. This gives $$k \int_{-\infty}^\infty e^{-ax^2} dx = f\left( \frac{a}{k^2} \right).$$ With a bit of rearrangement, we have the functional equation $$f(a) = \frac{1}{k} f\left( \frac{a}{k^2} \right)$$ whose unique solution on $(0, \infty)$ is $$f(a) = \frac{f(1)}{\sqrt a}.$$ This completes the proof without making any use of dimensional constraints, and is easily seen as equivalent to the dimensional analysis method demonstrated above.