Since you mention Laplace transforms, in its current form $\Gamma(s)$ is the Mellin transform of $e^{-x}$.
Here is another reason which is perhaps the most convincing. The Haar measure of a subset $S\subset \mathbb{R}^\times$ of the multiplicative group of real numbers is $\int_{x\in S} \frac{dt}{t}$, so the measure $\frac{dt}{t}$ over the real line is natural. The Gamma function is an analogue of a Gauss sum, and is the integral of multiplicative function $x^s$ against the additive function $e^{-x}$ over the measure of the group.
This problem was posed on Math Overflow, and received a large number of upvotes there. Take a look at the answers appearing on this thread: https://mathoverflow.net/questions/20960/why-is-the-gamma-function-shifted-from-the-factorial-by-1
You are right. The integral
$$\Gamma \left( x \right) = \int\limits_0^\infty {s^{x - 1} e^{ - s} ds},\,\,x\in\mathbb{R}\tag{1}$$
converges only if $x>0$. Therefore the definition only works for positive $x$. However, one can use the functional equation
$$\Gamma(x+1)=x\Gamma(x)$$
to give a meaning to $\Gamma(x)$ also when $x$ is negative. Namely, suppose $-1<x<0$. Then $x+1>0$, so $\Gamma(x+1)$ is defined by $(1)$. Now we define $\Gamma(x)$ (for which the formula $(1)$ doesn't make sense) as
$$\Gamma(x):=\frac{\Gamma(x+1)}{x}\tag{2}$$
You can already see that this definition makes no sense for $x=0$.
In the same way we can define $\Gamma(x)$ for all $x\in\mathbb{R}$ (except $0,-1,-2,\dots$)
Namely, if $-2<x<-1$, then $-1<x+1<0$, so we already know what $\Gamma(x+1)$ is by $(2)$, hence
$$\Gamma(x):=\frac{\Gamma(x+1)}{x}=\frac{\Gamma(x+2)}{x(x+1)}$$
In general, if $x>-n$, then
$$\Gamma(x):=\frac{\Gamma(x+n)}{x(x+1)\cdots(x-n+1)}$$
So you were interested in $\Gamma(-1/2)$. We get,
$$\Gamma(-1/2)=-2\Gamma(1/2)$$
which you can now calculate using $(1)$.
Best Answer
The gamma function does not satisfy any algebraic differential equation . But it is the solution of the following nonalgebraic differential equation: $$\frac{\partial w(x)}{\partial x}=w(x)~\psi(x);\qquad w(x)=\Gamma(x)$$
Otto Hölder proved in $1887$ that,
by showing that a solution to such an equation could not satisfy the gamma function's recurrence formula, making it a transcendentally transcendental function. This result is known as Hölder's theorem.