Why is signum function used to calculate Fourier transform of unit step function
If somebody you trust told you that the Fourier transform of the sign function is given by

$$\mathcal{F}\{\text{sgn}(t)\}=\frac{2}{j\omega}\tag{1}$$

you could of course use this information to compute the Fourier transform of the unit step $u(t)$. Using

$$u(t)=\frac12(1+\text{sgn}(t))\tag{2}$$

(as pointed out by Peter K. in a comment), you get

$$\mathcal{F}\{u(t)\}=\frac12\left(\mathcal{F}\{1\}+\mathcal{F}\{\text{sgn}(t)\}\right)=\pi\delta(\omega)+\frac{1}{j\omega}\tag{3}$$

However, you don't need the sign function to compute the Fourier transform of the step function. As suggested in your question, using the function $e^{-at}u(t)$ and taking the limit $a\rightarrow 0^+$ will also result in the expression given in $(3)$.

You can see this as follows. The Fourier transform of $e^{-at}u(t)$, $a&gt;0$, is given by

$$\int_0^{\infty}e^{-at}e^{-j\omega t}dt=\frac{1}{a+j\omega}\tag{4}$$

Taking the limit $a\rightarrow 0^+$ appears to give $1/j\omega$, but this is only valid for $\omega\neq 0$. Splitting the result $(4)$ in its real and imaginary part gives

$$\frac{1}{a+j\omega}=\frac{a}{a^2+\omega^2}+\frac{\omega}{j(a^2+\omega^2)}\tag{5}$$

The real part of $(5)$ is known as ($\pi$ times) a nascent delta function. It has the same form as the Poisson kernel, which in the limit becomes a Dirac delta impulse. So for $a\rightarrow 0^+$ the limit of $(5)$, and hence of $(4)$, is actually given by

$$\lim_{a\rightarrow 0^+}\frac{1}{a+j\omega}=\pi\delta(\omega)+\frac{1}{j\omega}\tag{6}$$

which equals the expression in $(3)$.