Solving the "Transport" PDE in the sense of distributions with Dirac Delta Source

The r.h.s. of the partially Fourier-transformed equation in OP is incorrect. Indeed, spatial Fourier transformation of the 2D Dirac $\delta_0 =\delta(x)\delta(t)$ gives $\delta(t)$, not $1$. Moreover, the weak form in OP is incorrect too. Integrating by parts, we have \begin{aligned} 0 &= \iint_{\Bbb R\times\Bbb R_+} (u_t + cu_x-\delta_0)\phi\,\text d x\,\text d t \\ &= -\int_{\Bbb R} g\phi|_{t=0}\, \text d x - \iint_{\Bbb R\times\Bbb R_+} u(\phi_t + c\phi_x)\,\text d x\,\text d t - \phi(0,0) \end{aligned} for any test function $\phi$.

The present problem amounts to the computation of the Green's function for the non-homogeneous advection equation $u_t+cu_x=f$. Fourier transformation in space and time of the PDE yields $$ -\text i(\omega-ck)\, \mathcal{F}_t\mathcal{F}_x u = 1 $$ where $\mathcal{F}_t = \int\text dt\, e^{\text i\omega t}$ and $\mathcal{F}_x = \int\text dx\, e^{-\text ik x}$. Thus, the solution is represented as \begin{aligned} u(x,t) &= \frac{1}{(2\pi)^2}\iint \frac{e^{-\text i(\omega t-kx)}}{\text i (kc-\omega)}\text dk\,\text d\omega \\ &= \frac{-1}{2\pi c}\int e^{-\text i\omega (t-x/c)}\text d\omega \\ &=-\tfrac{1}{c}\delta(t-x/c) \\ &=\delta(x-ct) \end{aligned} where the residue theorem was used (singularity at $k=\omega/c$). Using the superposition principle, the solution to the initial problem may be expressed as $$ u(x,t) = g(x-ct)+\delta(x-ct) \, . $$ Please let me know if you spot any mistake. One way to verify this result would be to evaluate the weak form of the PDE.

As pointed out in the comments, an alternative consists in using Duhamel's principle, cf. this article.

OP's first-order initial value problem (IVP) is

$$ \frac{\partial u(x,t)}{\partial t}+ c\frac{\partial u(x,t)}{\partial x}~=~\delta(t)\delta(x), \qquad u(x,t\!=\!0)~=~g(x).\tag{1}$$

One idea is to transform the IVP (1) into the form

$$ \frac{\partial v(x^{\prime},t^{\prime})}{\partial t^{\prime}}~=~\delta(t^{\prime})\delta(x^{\prime}), \qquad v(x^{\prime},t^{\prime}\!=\!0)~=~g(x^{\prime}),\tag{2}$$

by make a suitable linear coordinate transformation $(x,t)\mapsto (x^{\prime},t^{\prime})$. A bit of thought using the chain rule reveals that the coordinate transformation $$ x~=~x^{\prime}+ct^{\prime}, \qquad t~=~t^{\prime}, \tag{3}$$ will do the job. The unique solution to the IVP (2) is evidently $$ v(x^{\prime},t^{\prime})~=~\frac{1}{2}{\rm sgn}(t^{\prime})\delta(x^{\prime})+ g(x^{\prime}). \tag{4}$$ Hence the unique solution to the original IVP (1) is $$ u(x,t)~=~\frac{1}{2}{\rm sgn}(t)\delta(x\!-\!ct)+ g(x\!-\!ct). \tag{5}$$