= Solution
First prove the <convolution theorem> with the stated negative-exponential <Fourier transform>. Assuming absolute integrability sufficient for <Fubini's theorem>, change variables $u=t-s$:
$$
\begin{aligned}
\widetilde h(k)&=\int_{\mathbb R}\int_{\mathbb R}e^{-ikt}f(s)g(t-s)\,ds\,dt\\
&=\int_{\mathbb R}f(s)e^{-iks}ds\int_{\mathbb R}g(u)e^{-iku}du
=\widetilde f(k)\widetilde g(k).
\end{aligned}
$$
For the differential equation, <integration by parts> gives
$$
\widetilde {y'}=[e^{-ikt}y]_{-\infty}^\infty+ik\widetilde y=ik\widetilde y,
\qquad \widetilde {y''}=ik\widetilde {y'}=-k^2\widetilde y.
$$
The boundary terms vanish by the assumed decay of $y,y'$. As with the convolution step, the transforms can first be justified for integrable functions and derivatives; the resulting kernel gives the desired decaying response. Thus
$$
\boxed{\widetilde y(k)=\frac{\widetilde f(k)}{(a+ik)(b+ik)}=\widetilde f(k)\widetilde G(k).}
$$
The <convolution theorem> and <Fourier inversion theorem> imply $y=f*G$.
For the inverse transform of $\widetilde G$, use
$$
\widetilde G(k)=\frac1{a-b}\left(\frac1{b+ik}-\frac1{a+ik}\right).
$$
The <poles> are at $ib,ia$. For $t>0$, close the inverse-transform contour above; the residues of $e^{ikt}\widetilde G(k)$ there are $-ie^{-bt}/(a-b)$ and $ie^{-at}/(a-b)$. The inverse-transform factor $1/(2\pi)$ makes the answer $i$ times their sum. For $t<0$, close below, enclosing no poles. The large arcs vanish: the complete rational factor is $O(|k|^{-2})$ and the exponential is bounded on the appropriate semicircle. Therefore the <causal Green function of an overdamped oscillator> is
$$
\boxed{G(t)=\begin{cases}0,&t<0,\\(e^{-bt}-e^{-at})/(a-b),&t\geq0.\end{cases}}
$$
It is continuous, with $G(0)=0$ and $G'(0^+)=1$, so $(D_t+a)(D_t+b)G=\delta(t)$. The response can also be written $y(t)=\int_{-\infty}^tG(t-s)f(s)ds$.
For the bilateral exponential forcing, split the real <integral> at zero:
$$
\widetilde f(k)=\int_{-\infty}^0e^{(1-ik)t}dt+\int_0^\infty e^{-(1+ik)t}dt
=\frac1{1-ik}+\frac1{1+ik}=\boxed{\frac2{1+k^2}}.
$$
Consequently
$$
\boxed{\widetilde y(k)=\frac2{(1+k^2)(a+ik)(b+ik)}.}
$$
For negative time the lower half-plane contains only $k=-i$. The <residue> of $e^{ikt}\widetilde y$ there is $ie^t/[(a+1)(b+1)]$. Clockwise closure makes the inverse transform $-i$ times that residue. For positive time the three upper <residues>, at $i,ia,ib$ respectively, are
$$
-\frac{ie^{-t}}{(a-1)(b-1)},\qquad
-\frac{2ie^{-at}}{(a^2-1)(a-b)},\qquad
\frac{2ie^{-bt}}{(b^2-1)(a-b)}.
$$
Multiplication of their sum by $i$ gives the <bilateral exponential forcing of an overdamped oscillator>:
$$
\boxed{y(t)=\begin{cases}
\dfrac{e^t}{(a+1)(b+1)},&t\leq0,\\
\dfrac{e^{-t}}{(a-1)(b-1)}+\dfrac2{a-b}\left(\dfrac{e^{-at}}{a^2-1}-\dfrac{e^{-bt}}{b^2-1}\right),&t\geq0.
\end{cases}}
$$
Both expressions have $y(0)=y'(0)=1/[(a+1)(b+1)]$. This verifies the matching that prevents an extra <Dirac delta function> at zero. Direct substitution gives the forcing on either side, and all terms decay at their corresponding infinities. A causal <Green function> does not imply zero response before $t=0$ when the forcing has already been acting at negative times. The assumptions $a,b\ne1$ avoid repeated poles; their limiting cases would contain polynomial times exponential terms.
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