Is there an easy way to translate a Fourier transform table from angular frequency $\omega$ to Hertz $f$? I have a table with transform operations, e.g. scaling:

\begin{equation} \tag{0} \label{0} x(at) \iff \frac{1}{a} X(\frac{\omega}{a})  \end{equation}

or frequency shifting: 

\begin{equation}\tag{1} \label{1} x(t)e^{jw_ot} \iff X(\omega - \omega_o) \end{equation}

I also have a table with transform pairs, for example:

\begin{equation}\tag{2} \label{2} e^{-a|t|} \iff \frac{2a}{a^2+\omega^2} \end{equation}

Now, I would like to use these tables but with hertz $f$ instead. Note that there are many more operations and transform pairs in the tables than I have mentioned. I hope those can be translated similary as the examples I have given. Is there an easy way to do this?

For instance, I have tried to find the inverse transformation of $X(f - f_o)$ using equation $\eqref{0}$ and $\eqref{1}$

\begin{equation} X(f - f_o) =^? X(\frac{\omega - \omega_o}{2\pi}) \iff //eq. \eqref{0}// \iff 2\pi \cdot x(2\pi t)\cdot e^{j2\pi w_ot} =   \end{equation}

\begin{equation}\tag{3} \label{3} = 2\pi \cdot x(2\pi t)\cdot e^{j4\pi^2 f_ot}     \end{equation}

but equation $\eqref{3}$ gives \begin{equation}\tag{4} \label{4} \frac{1}{2}(X(f - f_o) + X(f + f_o)) \iff  2\pi \cdot x(2\pi t)\cdot \frac{e^{j4\pi^2 f_ot} + e^{-j4\pi^2 f_ot}}{2} \end{equation}

which simplifies to

\begin{equation}\tag{5} \label{5} \frac{1}{2}(X(f - f_o) + X(f + f_o)) \iff 2\pi \cdot x(2\pi t)\cdot \cos(j4\pi^2 f_ot) \end{equation}

which doesn't equal the correct expression

\begin{equation}\tag{6} \label{6} \frac{1}{2}(X(f - f_o) + X(f + f_o))  \iff x(t)\cdot \cos(j2\pi f_ot) \end{equation}.

Even if equation $\eqref{5}$ is correct, there is non-trivial calculation involved. I would like a faster way similar to "replace $\omega$ with $2\pi f$"

Thanks
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.