Is the exponential function the sole solution to these equations?

The answer is yes. Let me address the last stated form of the question. Functions $F$ with $F^{(n)}\geq 0$ are called totally monotone, and S. Bernstein proved that all such functions are Laplace transforms of positive measures, that is $$F(x)=\int e^{xt}d\mu(t).$$ Your functional equation $F(x+1)=eF(x)$ then implies $$\int e^{xt}(e^td\mu(t))=\int e^{xt}(ed\mu(t)).$$ From this follows $e^td\mu(t)=ed\mu(t)$ by the uniqueness theorem for the Laplace transform. This means that $\mu$ is an atom at the point $1$. This means that $F(x)=ce^x$, and your normalization implies $c=1$.

The references for Laplace transform are Widder, Laplace transform, and Feller, An introduction to probability... vol. II. Widder has Bernstein's theorem and uniqueness theorem.

Here's a proof that also uses Bernstein's representation theorem for totally monotone functions (I wondered if that result was needed, and Alexander Eremenko's answer now makes me think, that it is).

If $F:\mathbb{R}_+\to\mathbb{R}_+$ satisfies $F(x+1)=eF(x)$, it is of the form $F(x)=\exp(x+h(x))$, with $h$ a $1$-periodic function. But a totally monotonic function is logarithmically convex (see below) and $x+h(x)$ can only be convex if $h$ is a constant (actually zero, if also $F(0)=1$ or $F(1)=e$).

$$*$$

Totally monotonic functions are logarithmically convex:

From $F(x)=\int_{\mathbb{R}_+}e^{\lambda x}d\mu(\lambda)$, differentiating under the integral sign and applying the Cauchy-Schwarz inequality $$F'(x)^2=\bigg(\int_{\mathbb{R}_+}\lambda e^{\lambda x}d\mu(\lambda)\bigg)^2=\bigg(\int_{\mathbb{R}_+}\lambda e^{\lambda x/2}\cdot e^{\lambda x/2}d\mu(\lambda)\bigg)^2\le$$$$ \int_{\mathbb{R}_+}\lambda^2 e^{\lambda x}d\mu(\lambda)\int_{\mathbb{R}_+}e^{\lambda x}d\mu(\lambda)=F''(x)F(x)\ . $$

Therefore $(F'/F)'\ge0,$ that is $\log F(x)$ is convex.