Is it true that the Laplace Transform of a real function with compact support is always entire? Is it true that the Laplace Transform of a real function with compact support is always entire (entire = complex derivative exists on the entire complex plane)?

One should assume that the real function in question is integrable, otherwise the Laplace transform isn't defined at $\lambda=0$. Here is a more general statement.
Claim. Suppose $(\Omega, \mu)$ is a finite measure space and for every $\omega\in\Omega$ we have an entire function $f_\omega$. Suppose further that the family $\{f_\omega\}$ is uniformly bounded on compact subsets of $\mathbb C$. Then the function
$$F(z)=\int_\Omega f_\omega(z)\,d\mu(\omega) \tag1$$ 
is entire.
Proof. The integral (1) converges for all $z\in\mathbb C$, being the integral of a bounded function over a finite measure space. Since $f_\omega'$ are also uniformly bounded on compact subsets, the family $\{f_\omega\} $ is uniformly Lipschitz on compact subsets. This property  passes to $F$. 
Let $T$ be a triangle in the complex plane. The integral $\int_T F(z)\,dz$ makes sense by the above, and can be computed with Fubini's theorem: 
$$
\int_T F(z)\,dz = \int_\Omega \left(\int_T f_\omega(z)\,d\mu(\omega)\right)dz  =\int_\Omega \left(\int_T f_\omega(z)\,dz\right) d\mu(\omega) =0
$$
By Morera's theorem, $F$ is holomorphic in $\mathbb C$. $\quad\Box$
In our situation, $\Omega$ is a finite interval in $\mathbb R$, $d\mu(\omega) = f(\omega)\,d\omega$, and $f_\omega(z) = e^{-\omega z} $. The assumptions in the claim are satisfied, and $F$ in (1) is the Laplace transform of $f$.
