To determine the distributions it is sufficient to have two (non-parallel) vectors in $S^1$ and the assumptions of the Hamburger moment problem.

**Claim 1:** Suppose $a\neq b$, $a,b \in \mathbb{R}$ and that $Y$ and $Z$, independent, satisfy the assumptions of the Hamburger moment problem. Then the distributions of $Y + aZ$ and $Y + bZ$ determines the distribution of $Y$ and $Z$.

*Proof of Claim 1:* If we know the distribution of $Y +a Z$ and $Y+bZ$ we can calculate their expectations $\mathbf{E}(Y+aZ)$ and $\mathbf{E}(Y+bZ)$. Subtracting the two quantities we obtain $(a-b) \mathbf{E}(Z)$ which gives $\mathbf{E}(Z)$. Using that $\mathbf{E}(Y) = \mathbf{E}(Y+aZ) - a \mathbf{E}(Z)$ we can uniquely identify the first moment of $Y$ and of $Z$. So far, no independence needed.

Using independence, we can also calculate the second moments, say $m_{a,2}$ and $m_{b,2}$:
\begin{align}
 m_{a,2} &= \mathbf{E}((Y+aZ)^2) = \mathbf{E}(Y^2) + 2a \mathbf{E}(Y) \mathbf{E}(Z) + a^2 \mathbf{E}(Z^2)  \, ,\\
 m_{b,2} &= \mathbf{E}((Y+bZ)^2) = \mathbf{E}(Y^2) + 2b \mathbf{E}(Y) \mathbf{E}(Z) + b^2 \mathbf{E}(Z^2) \, ,
\end{align}
Again subtracting the two equations lets you obtain:
\begin{equation}
 (a^2 -b^2) \mathbf{E}(Z^2) = m_{a,2}-m_{b,2} -2 (a+b) \mathbf{E}(Y) \mathbf{E}(Z) \, .
\end{equation}
This gives you $\mathbf{E}(Z^2)$ and $\mathbf{E}(Y^2)$ as before. Iterating this procedure we can construct $\mathbf{E}(Z^k)$ and $\mathbf{E}(Y^k)$ for any $k\in \mathbb{N}$. Supposing that $Y$ and $Z$ satify certain conditions (see Hamburger moment problem), this uniquely defines the distribution of $Y$ and $Z$.q.e.d.

**Claim 2:** Suppose $(V_1,v_2), (w_1,w_2) \in S^1$ with $v_1 w_1 + v_2 w_2 \in (-1,1)$ (not parallel) and that $X_1, X_2 \in \mathbb{R}$, independent, satisfy the assumptions of the Hamburger moment problem. Then the distributions of $v_1 X_1 + v_2 X_2$ and $w_1 X_1 + w_2 X_2$ determine the distribution of $Y$ and $Z$.

*Proof of Claim 2:* 
If $v_1 = 0$, this means we know the distribution of $v_2 X_2$, i.e. that of $X_2$. Since $|v_2|=1$ we know that $|w_2|<1$ and thus $w_1 \neq 0$ allowing us to reconstruct $X_1$ from the distributions of $w_1 X_1 + w_2 X_2$ and $X_2$ by independence$^{1)}$. If $w_1 = 0$, likewise.

Suppose now that $v_1 w_1 \neq 0$. Then we know the distribution of $X_1+ \frac{v_2}{v_1} X_2$ and that of $X_1 + \frac{w_2}{w_1}X_2$. Using Claim 1 allows to determine the distributions of $X_1$ and $X_2$ uniquely. q.e.d.

**Regarding the moment problem:** Example 6.4 in an article by Svante Janson, http://arxiv.org/abs/math/0605642, tells you that it might be difficult to have the result holding without moment assumptions. However, he has no independence assumptions, so maybe one cannot easily find a counterexample in your case.

***For completeness, one point is not enough:*** It will in general not suffice to know $Y+aZ$ for a single value of $a$ (even if $\neq 0$) to determine the distributions of (independent) $Y$ and $Z$: consider $Y, Z \sim N(0,1)$ and $Y' \sim N(0,1+a^2/2), Z' \sim N(0,1/2)$, both independent.

$1)$ Not having independence might require the moment problem conditions again.