Sketch: We want to prove that for any $p\in S$ and $v\in T_pS$, the second derivative of $F$ at $p$ in the direction $v$ is positive. Note that for any given such $p,v,$ everything that matters to us happens in the $2$-dimensional space $\mathrm{span}(p,v).$ Hence, it is enough to solve the problem for $n=1$.
So, we have a convex curve $S$ in $\mathbb{R}^2$, and we have some $p\in S$. Let $$l=\{p+t\cdot v|t\in\mathbb{R}\}$$ denote the line tangent to $S$ at $p$. (In other words, $0\neq v\in T_pS$). By definition of $F$, we have $F(p)=1$. By convexity, we have $F(q)>1$ for every $p\neq q\in l$. This means that the function $F|_l$ admits a minimum at $p$, and hence, the second derivative in the direction $v$ is positive. (Well, non-negative, anyway. All you have to do now is use strict convexity in order to show that the second derivative does not vanish at $p$).
Edit: We show that strong convexity of a hypersurface is independent of the defining function. Let $S\subset\mathbb{R}^n$ be defined locally both by $\{f=0\}$ and $\{g=0\}$, where $f,g:U\to\mathbb{R}$ are smooth and both derivatives $df, dg$, don't vanish on $S$. Let $p\in S\cap U$. It follows that there is a smooth $h:U\to\mathbb{R}$, such that $g=hf$. Since $dg_p\neq0,$ it follows from the Leibniz rule that necessarily $h(p)\neq0$.
We assume now that $S$ is strongly convex with respect to $f$, that is, $\nabla^2f_p$ is positive definite on the tangent space $T_pS$. Let $v\in T_pS$. Then$$dg(v)=dh(v)f+hdf(v),$$and at $p$ we have$$\nabla^2g_p(v)=\nabla^2h_p(v)f(p)+2dh_p(v)df_p(v)+h(p)\nabla^2f_p(v).$$By the assumptions, the first and second terms vanish. The last term has the sign of $h(p)$. Thus, if $g$ is such that $h$ is positive, then $\nabla^2g_p$ is positive definite on $T_pS$, and if $h$ is negative, $\nabla^2g_p$ is negative definite on the tangent space.
Edit 2: The above argument works whenever $f$ and $g$ are at least $C^3$. We now treat the case where they are only $C^2$ (then $h$ is $C^2$ far away from $S$, but only $C^1$ on $S$). Since $g$ is $C^2$, we have$$\nabla^2g_p(v)=\lim_{x\to p}\nabla^2g_x(v)=\lim_{x\to p}\nabla^2h_x(v)f(x)+2dh_x(v)df_x(v)+h(x)\nabla^2f_x(v),$$ where we let $x$ approach $p$ from outside of $S$, and so the limit is well defined. It suffices then to prove$$\lim_{x\to p}\nabla^2h_x(v)f(x)=0.$$For that, we change the coordinates and assume $S=\{x_n=0\}.$ Note that when we do so, the expression for the Hessian tensor includes a correction involving the Christoffel symbols. However, this correction is bounded, as it involves only the first derivatives of $h$, and hence vanishes when multiplied by $f$, as $x$ gets closer to $p$. Consequently, we may consider the standard Hessian in our new coordinates.
Outside of $S$, we have $h=g/f$. Hence, for $1\leq i\leq n-1$,$$h_i=\frac{g_if-gf_i}{f^2}.$$Differentiating again,
\begin{align}
h_{ii}&=\frac{(g_{ii}f+g_if_i-g_if_i-gf_{ii})f^2-(g_if-gf_i)2ff_i}{f^4}\\
&=\frac{g_{ii}}{f}-\frac{gf_{ii}}{f^2}-\frac{2g_if_i}{f^2}+\frac{2gf_i^2}{f^3},
\end{align}
and hence,
\begin{align}
h_{ii}f&=g_{ii}-\frac{g}{f}f_{ii}-2f_i\frac{g_if-gf_i}{f^2}\\
&=g_{ii}-hf_{ii}-2f_ih_i.
\end{align}
Since on $S$ both $f$ and $g$ are constant, all the terms vanish on $S$. It follows that$$\lim_{x\to p}\nabla^2h_x(v)f(x)=0,$$and the proof is complete.
Best Answer
Since nobody closed or answered this, I will self-answer and self-accept, so as to make it answered.
That the Hessian be everywhere p.d. implies the gradient is injective, as proved here. Basically, the proof uses Taylor expansion with integral form for the remainder, then using p.d.-ness to prove the remainder is always strictly positive.
$f(x)=e^x$ has everywhere p.d. Hessian (i.e. everywhere strictly positive second derivative) but is surely not surjective.