Your trick doesn't really work. There are several issues.
First, your trick shows that for every unsatisfiable circuit of size $n$, there exists a satisfiable circuit of size $n+1$. But that doesn't tell you anything about the relationship between the number of unsatisfiable circuits of size $n$ vs the number of satisfiable circuits of size $n$. And, in general, there are way more circuits of size $n+1$ than circuits of size $n$, and most circuits are satisfiable, so the mapping you've found doesn't prove what you want it to prove. To prove what you're trying to prove, you'd need to find a bijection between unsatisfiable circuits of size $n$ and satisfiable circuits of size $n$ (not $n+1$); no such bijection exists.
Second, the mapping you describe isn't a bijection. Consider for example the circuit $\phi(x) = x$. This is satisfiable. Slapping a "not" in front if it, to get $\phi'(x) = \neg x$, yields another circuit that is also satisfiable. Many circuits are both satisfiable (there exists an assignment that makes $\phi$ true) and their-complement-is-satisfiable (there exists a different assignment that makes $\phi$ false). So negating an unsatisfiable circuit makes it satisfiable, but negating a satisfiable circuit doesn't necessarily make it unsatisfiable.
Finally, there's the distinction between circuits vs formulas, as David Richerby explains. If you negate the output of a CNF formula, the result isn't a CNF formula.