**Fact**: if you can recover $(f,g)$ from the collection $(h_v)_{v\in V}$ then it must be the *unique* solution to your problem.

The case $h_v=0$ leads to a lot of non-uniqueness. Otherwise, there is only one source of non-uniqueness: the existence of functions $\alpha,\beta$ solving $$ \alpha(v_1s)\beta(v_2s)=1 .$$

If $\alpha,\beta$ are continuous then they must be constant as long as $V$ contains two non-colinear vectors.

**Answer**: Yes under some additionnal hypothesis. An important one is $f(0)g(0)\neq0$, so let us assume that. Assume also that $f,g$ are differentiable at $0$ and that $V$ contains two non-colinear vectors $v, \tilde v$ (this hypothesis is sufficient), and define $$\tau:=\frac{v_1\tilde v_2}{v_2\tilde v_1}.$$ We can label $v,\tilde v$ in such a way that $0\leq|\tau|<1$ and $v_2\neq0$.

Fiddling around with the functional equation gives you, for $t\in \mathbb R$ with $f(\frac{v_1 }{ v_2}t)\neq0$: $$ g(t)=g(\tau t)\phi(t) ,$$ where $$\phi(t):=\frac{h_v(\frac{1}{v_2}t)}{h_{\tilde v}(\frac{v_1 }{\tilde v_1 v_2}t)}.$$ Now you can iterate: $$g(t)=g(0)\prod_{n=0}^{\infty}\phi(\tau^n t).$$ Because $\phi(0)=1$ and is differentiable at $0$, there exists a constant $C$ such that, for $n$ big enough, $$|\phi(\tau ^n t)-1|\leq C|\tau^n t|.$$ The infinite product converges absolutely (and locally uniformly in $t$), so that $g$ is computed in terms of $h_v$, $h_{\tilde v}$ and $g(0)$ only. You can recover $f$ easily from the functional equation.