Inverse Fourier Transform of $ (j\omega)\cdot \frac{1}{a+j\omega} $ We are trying to find the inverse Fourier Transform of $$ (j\omega) \cdot \frac{1}{(a+j\omega)} $$
For this we would obviously need to use Fourier transform pairs / equations
$ e^{-at} \cdot u(t), a > 0 \xrightarrow{\text{X(j$\omega$)}} \frac{1}{(a+j\omega)} \tag 1$ 
$ t \xrightarrow{\text{X(j$\omega$)}} j\omega \tag 2$
However, we disagree that equation (2) holds true for this situation. If it does, this would then yield $x(t) = t \cdot e^{-at} \cdot u(t) $. We have difficulty understanding if htis would hold applicable or not: any helpful thoughts / suggestions?

Let's start by stating the convolution theorem:
$$\text{For $f,g$ functions we have: } \begin{align}\mathcal F(f\cdot g)=\mathcal F(f)*\mathcal F(g)\\\mathcal F(f* g)=\mathcal F(f)\cdot\mathcal F(g)\end{align}$$
This can be extended to the inverse easily:
$$\mathcal F^{-1}=R\mathcal F=\mathcal FR;\text{ where $R$ is flip operator}$$
Hence
$$\text{For $f,g$ functions we have: } \begin{align}\mathcal F^{-1}(f\cdot g)=\mathcal F(Rf\cdot Rg)=\mathcal F(Rf)*\mathcal F(Rg)=\mathcal F^{-1}(f)*\mathcal F^{-1}(g)\\\mathcal F^{-1}(f* g)=\mathcal F(Rf* Rg)=\mathcal F(Rf)\cdot\mathcal F(Rg)=\mathcal F^{-1}(f)\cdot\mathcal F^{-1}(g)\end{align}$$
We have $2$ functions: $f=j\omega, g=\frac{1}{a+j\omega}$
We can calculate to see that $\mathcal F^{-1}(f)=\mathcal F(Rf)=\mathcal F((-jw)^1\cdot1)=\delta'(t)$(I don't know where you got your value but it was wrong)
Unless I did a stupid mistake in my calculation, the inverse fourier transform of $g$ is exactly what you got.
Because we take the inverse of the product we change it to the convolution of the inverse:
$$\mathcal F^{-1}\left(j\omega\cdot\frac1{a+j\omega}\right)=\left(\delta'(t)*(e^{-at} \cdot u(t))\right)=\dfrac d{dt}\left(e^{-at} \cdot u(t)\right)=\delta(t)-au(t)e^{-at}$$(assuming, ofc, that $a>0$)

Edit
It is important to point out, this is without normalization. Because I don't know from where the function you have got I can't tell if we need normalization, and if yes which one. The $2$ most common normalizations are: dividing by $2\pi$ only at the inverse or dividing by $\sqrt{2\pi}$ both on the forward and at the inverse fourier transform
