= Solution
Let $M=\sup|f|$ and $F_T(z)=\int_0^T f(t)e^{-tz}dt$. Each $F_T$ is an <entire function>: on any compact set of $z$, differentiation under this finite-interval integral is justified by a bounded integrable majorant. We prove $\boxed{F_T(0)\to F(0)}$, which gives ordinary convergence of the improper integral, not just an Abel limit.
Fix $R>0$. Since $\Omega$ contains the imaginary segment from $-iR$ to $iR$ and is open, choose $0<\delta<R$ so small that its closed left strip of width $\delta$ belongs to $\Omega$. Let $C_+$ be the right semicircle $|z|=R$ oriented from $-iR$ to $iR$, and let $\Gamma$ be the left half-ellipse $z=\delta\cos\theta+iR\sin\theta$, $\pi/2\le\theta\le3\pi/2$, oriented back to $-iR$. The enclosed right half-disk and thin left region lie in $\Omega$. The <residue theorem>, applied to the analytic difference with its simple pole at zero, gives
$$
F_T(0)-F(0)=\frac1{2\pi i}\int_{C_+\cup\Gamma}(F_T(z)-F(z))e^{Tz}\left(1+\frac{z^2}{R^2}\right)\frac{dz}{z}.
$$
For $x=\operatorname{Re}z>0$,
$$
|F_T(z)-F(z)|\le\int_T^\infty M e^{-tx}dt=\frac{M e^{-Tx}}x.
$$
On the right circle, $|1+z^2/R^2|=2x/R$ and $|z|=R$. The exponential cancels the displayed damping, so the absolute integrand per unit arc length is at most $2M/R^2$. The right-arc contribution is therefore at most $M/R$ after the factor $1/(2\pi)$.
For the $F_T$ term on $\Gamma$, deform that left path to the left semicircle $C_-$ of radius $R$. The <Cauchy integral theorem> allows this because $F_T$ is an <entire function> and the region between the paths excludes the pole at zero. On $C_-$, $x<0$ and
$$
|e^{Tz}F_T(z)|=\left|\int_0^T f(t)e^{(T-t)z}dt\right|\le\frac{M}{-x}.
$$
The same factor $|1+z^2/R^2|=2|x|/R$ bounds this contribution by $M/R$. The arc endpoints can be omitted from the estimates, since they have zero arc length and the integrands are continuous there.
Finally, the $F$ term on the original $\Gamma$ tends to zero as $T\to\infty$. This compact path stays away from zero, so $F(z)(1+z^2/R^2)/z$ is bounded there. Its interior has $\operatorname{Re}z<0$, hence $e^{Tz}\to0$, and $|e^{Tz}|\le1$. <Dominated convergence> on the finite-length path gives the claim. Combining the three pieces yields
$$
\limsup_{T\to\infty}|F_T(0)-F(0)|\le\frac{2M}{R}.
$$
Since $R$ is arbitrary, the difference tends to zero. This proves the <Newman Tauberian theorem> in the required setting; the proof crucially uses analytic continuation across every finite segment of the imaginary axis.
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