Farkas' Lemma is indeed the way to go, but we need the right setting. Below I give a sketch.
For simplicity, assume that we work at a vertex $x=0$ of $P$.
So we want to find a minimal set of generators for the cone $\DeclareMathOperator{\cone}{cone}C:=\cone(P)=\cone (\mathcal V)$, where $\mathcal V\subseteq P$ is the set of vertices of $P$.
What we want to understand is whether every such "minimal generator" $y\in\mathcal V$ is a neighbor of $x$, because if so, then the edge-directions indeed generate $C$.
So, suppose that $y\in \mathcal V$ is part of such a minimal set of generators. Then $y\not\in C':=\cone(\mathcal V\setminus \{y\})$ (here you need to use that no three vertices of $P$ are colinear). By Farkas' Lemma, we can then separate $y$ from $C'$ via a hyperplane. In particular, we can choose this hyperplane with normal vector $n$ so that
$$\def\<{\langle}\def\>{\rangle}\<n,x\>=0,\quad\<n,y\> >0\quad\text{and}\quad\<n,z\><0\text{ for all $z\in \mathcal V\setminus\{x,y\}$}.$$
It is not too hard to argue that we can choose $n$ linearly independent from $y$ (if we are working in dimension $d\ge 2$). Then
$$n':=n-y\frac{\<n,y\>}{\<y,y\>} \not=0.$$
You can check that we have $\<n',x\>=\<n',y\>=0$ and $\<n',z\><0$ for all $z\in \mathcal V\setminus\{x,y\}$ (the latter needs some thought, but is possible). In other words, the hyperplane orthogonal to $n'$ supports $P$ exacty at the two vertices $x$ and $y$, which proves that these form an edge of $P$.
In still other words, $\cone(P)$ is generated by the neighbors of $x$.
Some further explanation
As requested in the comments, I elaborate on $\<n',z\><0$ for all $z\in\mathcal V\setminus\{x,y\}$. As Epiousios noted, this is the same as
$$(*)\quad \underbrace{\<n,z\>}_{<0} < \underbrace{\frac{\<n,y\>}{\<y,y\>}}_{>0} \<y,z\>,$$
which would be obviously true if $\<y,z\>>0$. However, this is not always the case.
But, we can do a trick: before we start with any of our argument, we can tranform our polytope $P$ into an more convenient polytope $P'$, for which any two neighbors $y,z$ of $x=0$ satisfy $\<y,z\>>0$ (meaning $\sphericalangle(y,z)<90^\circ$).
We can do this by stretching $P$ in a certain way. Hopefully, the following image makes this clearer:
Since this is a linear transformation, this changes nothing about the actual problem. But this time $(*)$ is trivially satified.
Best Answer
Let me say something about the case for polyhedra (it may generalize, but I am not sure about the details).
It is not too hard to imagine that there might be a "combinatorial duality" for spherical polyhedra, in the sense that the dual of a spherical polyhedron exists, but is only determined up to combinatorial equivalence (e.g. via dual planar graphs). But I want to argue that there can not be a geometric duality, i.e. a duality that to every concrete spherical polyhedron gives you another one, and taking the dual again brings you back to the original one.
The reason is, that given the combinatorial type of a spherical polyedron, the realization space of that type (i.e. the space of all spherical polyhedra with this combinatorial type) has a local dimension of $2n$, where $n$ is the number of vertices.
What do I mean by that: you can describe your spherical polyhedron basically by drawing some points on the sphere, and stating between which points there should be a line. The line is then uniquely determined as the great circle arc between these points (yes, there is a choice which arc to take, but lets ignore this for now). So if we placed our points carefully, then none of these arcs intersect, and what we have is a spherical polyhedron.
But note that we can move each point slighly, and the arcs move accordingly. And if we moved the points slightly enough, then the arcs stay disjoint, and the construct stays a spherical polyhedron. Since each vertex moves on the surface of the 2-sphere, each vertex has two degrees of freedom, and the whole construct has $2n$ degrees of freedom.
Now, consider the spherical cube, whose dual (if our duality is meaningful in any way) is the spherical octahedron. But the first one has $2\times 8=16$ degrees of freedom, and the latter one only $2\times 6=12$. So not every unique realization of the spherical cube can be mapped into a unique realization of the spherical octahedron, and so the geometric duality fails.