= Solution
For $t>0$, use the continuous logarithm of the <characteristic function>, normalized to zero at $u=0$, to write
$$
\frac1t\log\phi_{X_t}(u)
=a(\cos(bu)-1)+c(e^{iu}-1)+icu.
$$
Thus the deterministic drift is \b[$+c$]. In particular the last contribution is not the negative drift of a compensated unit-jump process.
Take a <Poisson random measure> $N$ on $(0,\infty)\times\{+,-,p\}$ with intensity
$$
ds\left(\frac a2\delta_++\frac a2\delta_-+c\delta_p\right).
$$
Define the jump marks $j(+)=b$, $j(-)=-b$, and $j(p)=1$. A realization with the required process law is the <Poisson stochastic integral with finite intensity>
$$
\boxed{X_t=ct+\int_{(0,t]\times\{+,-,p\}}j(z)\,N(ds,dz)
=ct+bN_t^+-bN_t^-+N_t^p,}
$$
where the three counts are independent <Poisson processes> of rates $a/2,a/2,c$. Their <characteristic functions> multiply to
$$
\exp\left\{t\left[icu+\frac a2(e^{ibu}-1)+\frac a2(e^{-ibu}-1)+c(e^{iu}-1)\right]\right\},
$$
which is exactly the given expression. The constructed process is a <Lévy process>, and stationary independent increments make its entire finite-dimensional law determined by these one-time characteristic functions. This gives a representation in law of the specified process.
Equivalently, using its nonzero-jump measure, the <Lévy measure> is
$$
\nu=\frac a2\mathbf1_{\{b>0\}}(\delta_b+\delta_{-b})+c\delta_1,
$$
and the unmarked representation is $X_t=ct+\int_{(0,t]\times(\mathbb R\setminus\{0\})}z\,N(ds,dz)$ with intensity $ds\,\nu(dz)$. The finite-jump case of the <Lévy–Itô decomposition> realizes this pathwise using the jump measure of a version of $X$; the integral is an uncompensated finite sum.
The <atomic compound Poisson process with drift> has <càdlàg> paths with finitely many nonzero jumps on every bounded time interval, linear slope $c$ between jumps, and no Brownian component. When $b>0$, jumps of sizes $b,-b,1$ occur at the stated rates; when $b=1$, positive unit-jump rates combine to $a/2+c$. When $b=0$, the two symmetric marks have zero effect and are omitted from the <Lévy measure>; the process reduces to $ct+N_t^p$, so $a$ has no effect. If $c=0$ the paths are piecewise constant, and if also $ab=0$ they are identically zero. All paths have finite variation on bounded intervals. As a check on the drift sign,
$$
\mathbb EX_t=2ct,\qquad\operatorname{Var}(X_t)=t(ab^2+c).
$$
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