Is there an easy way to translate a Fourier transform table from angular frequency $\omega$ to Hertz $f$?
Your confusion comes from the fact that you use $X(\cdot)$ for denoting both functions, the function of $\omega$ and the function of $f$, but they are really two different functions, because

$$X(\omega)=\hat{X}(f)=\int_{-\infty}^{\infty}x(t)e^{-j2\pi ft}dt,\quad\omega=2\pi f\tag{1}$$

That's why in all the correspondences you mentioned you can just replace $\omega$ by $f$ inside the function argument (and note that the two frequency domain functions are actually different).

Note, however, that outside the function you have to replace $\omega$ by $2\pi f$, as for instance in the following correspondence:

$$\frac{d^nx(t)}{dt^n}\Longleftrightarrow (j\omega)^nX(\omega)=(j2\pi f)^n\hat{X}(f)\tag{2}$$

There is one more complication that arises due to the transform as defined by $(1)$ being unitary if $\hat{X}(f)$ is used, but being non-unitary when using $X(\omega)$. This can be seen by the factor $1/(2\pi)$ for the inverse transform when using $X(\omega)$. This implies, for instance, that you need a factor $1/(2\pi)$ in the following correspondence when using $X(\omega)$, but not when you use $\hat{X}(f)$:

$$x(t)y(t)\Longleftarrow \frac{1}{2\pi}(X\star Y)(\omega)=(\hat{X}\star\hat{Y})(f)\tag{3}$$

Also be careful when a Dirac delta impulse is involved, because clearly $\delta(\omega)\neq\delta(f)$. The correct correspondence is

$$\delta(f)=2\pi\delta(\omega)\tag{2}$$

Consequently,

$$\mathcal{F}\{1\}=2\pi\delta(\omega)=\delta(f)\tag{3}$$

and

$$\mathcal{F}\{e^{j\omega_0t}\}=\mathcal{F}\{e^{j2\pi f_0t}\}=2\pi \delta(\omega-\omega_0)=\delta(f-f_0)\tag{4}$$

from which the Fourier transforms of sine and cosine follow easily.