I'll try to rearrange the problem: we want to calculate the difference of potential $y(t)$ using a Fourier transform and the convolution theorem. We can now write the Kirchoff equation for the circuit:
$$y(t)=U(t)-L\frac{dI}{dt}(t)$$
and then we can apply the Fourier transform on the two sides of the equation obtaining this (using the tilde symbol for the transform):
$$\tilde{Y}(\omega)=\tilde{U}(\omega)-i\omega L \tilde{I}(\omega)$$
Here we have applied the Fourier rule for the derivative of a function. Note that now we have a simple equation in $\tilde{I}$ since we can write $\tilde{Y}(\omega)=R\tilde{I}(\omega)$. Solving the equation for $\tilde{I}$ leads to a simple expression for $\tilde{Y}(\omega)$:
$$\tilde{Y}(\omega)=\frac{R\tilde{U}(\omega)}{R+i\omega L }$$
We can now use the convolution theorem and split out the previous equation in two terms $h(t)$ and $g(t)$ in order to have $\tilde{Y}(\omega)=F[h(t)*g(t)]=\sqrt{2\pi}F[h(t)]F[g(t)]=\sqrt{2\pi}\tilde{H}(\omega)\tilde{G}(\omega)$:
$$\tilde{Y}(\omega)=\sqrt{2\pi}\left[\frac{1}{\sqrt{2\pi}}\frac{1}{R+i\omega L}\right][R\tilde{U}(\omega)]$$
If we calculate the antitrasform of the two functions in the brackets, then we have a convolution integral for $y(t)$. We know that the antitrasform of $R\tilde{U}(\omega)$ will be $RU(t)$, so we can only calculate one antitrasform:
$$F^{-1}[\tilde{H}(\omega)]=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{e^{i\omega t}}{R+i\omega L}$$
Now we must choose a suitable closed path of integration on which we can apply the Cauchy theorem. We can for example use the half circumference in the Argand-Gauss plan that lies on the positive part of the ordinates axis and then complete the path with a straight line on the abscissa's axis. Inside this integration path we have a pole of order 1 (namely $\omega=i\frac{R}{L}$). Using the Jordan's lemma we can say that the integral over the circumference is 0 and so we only have to calculate the residue in $\omega=i\frac{R}{L}$. Finally we conclude that
$$y(t)=\frac{1}{\tau}\int_{-\infty}^{+\infty} e^{-(t-t')/\tau}U(t')dt'$$
where $\tau=L/R$. This is the convolution of a given input signal $U(t)$ and the function $G(t)=\frac{1}{\tau}e^{-t/\tau}$. In particular $G(t)$ is referred to as the system's transfer function or gain. What's the meaning of $G(t)$? It represents the impulsive response of the system because we have $y(t)=G(t)$ for a delta-like input $U(t)=\delta(t)$. What's the meaning of the convolution? If we have many inputs that follow one another, the overall output must be the sum of the responses of the systems, in fact, linear systems like RL circuits obey to the superposition principle. For example, if we have two delta-like inputs that follow one another, we must sum the two exponential responses to obtain $y(t)$. Note that in this case, if the delta inputs are near in time (the time interval must be $<\tau$), the maximum value for $y(t)$ is reached when the second delta impulse is applied to the system.
