Suppose that a co-finite subset of the $a_p$'s generate a field $L$. Applying any element $\sigma$ of $\mathrm{Gal}(K_f/L)$ to $f$ yields another form $f'$ whose Fourier coefficients $b_p$ are equal to $a_p$ for all but finitely many $p$. By strong multiplicity one, $f'=f$ and $b_p = a_p$ for all $p$. Therefore $\sigma$ is trivial and $L = K_f$.

Suppose that a co-finite subset of the $a_p$'s generate a field $L$. Applying any element $\sigma$ of the absolute Galois group of $L$ to $f$ yields another form $g$ whose Fourier coefficients $b_p$ are equal to $a_p$ for all but finitely many $p$. By strong multiplicity one, $g=f$ and $b_p = a_p$ for all $p$. Therefore $\sigma$ fixes $K_f$ as well, and since $L \subset K_f$ it follows that $L=K_f$.