To determine the distributions it is sufficient to have two (non-parallel) vectors in $S^1$, with the first coordinate not having the same absolute value, and the assumptions of the Hamburger moment problem.
Claim 1: Suppose $a\neq b$, $a,b \in \mathbb{R}$ and that $Y$ and$(v_1,v_2), (w_1,w_2) \in S^1$ with $Z$, independent, satisfy the assumptions of the Hamburger moment problem$v_1^2 \neq w_1^2$. Then the distributions of $Y + aZ$ and $Y + bZ$ determines the distribution of $Y$ and$v_1^k w_2^k - v_2^k w_1^k \neq 0$ for any $Z$$k \in \mathbb{N}$.
Proof of Claim 1: If we know the distribution of $Y +a Z$ andProof: Assume $Y+bZ$ we can calculate their expectations$v_1^k w_2^k = v_2^k w_1^k$ for some $\mathbf{E}(Y+aZ)$$k \in \mathbb{N}$ and $\mathbf{E}(Y+bZ)$. Subtracting the two quantities we obtain $(a-b) \mathbf{E}(Z)$ which gives $\mathbf{E}(Z)$lead this to a contradiction. Using thatIf $\mathbf{E}(Y) = \mathbf{E}(Y+aZ) - a \mathbf{E}(Z)$$k$ is even, we can uniquely identifytake the first moment of $Y$ and of$k/2$-th root to obtain $Z$. So far, no independence needed$v_1^2 w_2^2 = v_2^2 w_1^2$.
Using independence, Now using that we can also calculateare on the second momentssphere, say $m_{a,2}$ and $m_{b,2}$: \begin{align} m_{a,2} &= \mathbf{E}((Y+aZ)^2) = \mathbf{E}(Y^2) + 2a \mathbf{E}(Y) \mathbf{E}(Z) + a^2 \mathbf{E}(Z^2) \, ,\\ m_{b,2} &= \mathbf{E}((Y+bZ)^2) = \mathbf{E}(Y^2) + 2b \mathbf{E}(Y) \mathbf{E}(Z) + b^2 \mathbf{E}(Z^2) \, , \end{align} Again subtracting the two equations lets you obtain: \begin{equation} (a^2 -b^2) \mathbf{E}(Z^2) = m_{a,2}-m_{b,2} -2 (a+b) \mathbf{E}(Y) \mathbf{E}(Z) \, . \end{equation} This gives you $\mathbf{E}(Z^2)$ and $\mathbf{E}(Y^2)$ as before. Iterating this procedure we can constructget $\mathbf{E}(Z^k)$$v_1^2(1-w_1^2) = (1-v_1^2)w_1^2$ and $\mathbf{E}(Y^k)$ for anythis is equivalent to $k\in \mathbb{N}$$v_1^2 = w_1^2$ wich is a contradiction to the assumption. Supposing that $Y$ and $Z$ satify certain conditionsIf (see Hamburger moment problem)$k$ is odd, this uniquely definesthen the distribution of $Y$ andassumption holds only if $Z$$v_1 w_2 = v_2 w_1$ which is a special case of the even case.q q.e.d.
Claim 2: Suppose $(V_1,v_2), (w_1,w_2) \in S^1$$(v_1,v_2), (w_1,w_2) \in S^1$ with $v_1 w_1 + v_2 w_2 \in (-1,1)$ (not parallel) and that$v_1^2 \neq w_1^2$. Let $X_1, X_2 \in \mathbb{R}$, independent, satisfy the assumptions of the Hamburger moment problem. Then the distributions of $v_1 X_1 + v_2 X_2$ and $w_1 X_1 + w_2 X_2$ determine the distributiondistributions of $Y$$X_1$ and $Z$$X_2$.
Proof of Claim 2: If $v_1 = 0$, this means we know the distribution of $v_2 X_2$, i.e. that of $X_2$. Since $|v_2|=1$ we know that $|w_2|<1$ and thus $w_1 \neq 0$ allowing usProof: We want to reconstruct $X_1$ fromdeduce the distributions of the moments of $w_1 X_1 + w_2 X_2$$X_1$ and $X_2$ by independence$^{1)}$iteratively. If $w_1 = 0$, likewiseThis suffices since both distributions satify the assumptions of the Hamburger moment problem.
Suppose now thatwe know the moments from order $v_1 w_1 \neq 0$$1$ to order $k-1$ for some $k \in \mathbb{N}$. ThenBy assumption we know the distribution of $$ Y:= v_1 X_1 + v_2 X_2 \, , \quad Z:= w_1 X_1 + w_2 X_2 \, .$$ Consider $X_1+ \frac{v_2}{v_1} X_2$$\mathbb{E}[Y^k]$ and that of $X_1 + \frac{w_2}{w_1}X_2$. Using$\mathbb{E}[Z^k]$: \begin{align} m_{Y,k} &= \mathbf{E}(Y^k) = v_1^k \mathbf{E}(X_1^k) + \sum_{l=1}^{k-1} {k\choose l} v_1^l v_2^{k-l} \mathbf{E}(X_1^{l}) \mathbf{E}(X_2^{k-l}) + v_2^k \mathbf{E}(X_2^k) \, ,\\ m_{Z,k} &= \mathbf{E}(Z^k) = w_1^k \mathbf{E}(X_1^k) + \sum_{l=1}^{k-1} {k\choose l} w_1^l w_2^{k-l} \mathbf{E}(X_1^{l}) \mathbf{E}(X_2^{k-l}) + w_2^k \mathbf{E}(X_2^k) \, , \end{align} Using Claim 1 allows to determinewe can eliminate the distributions$\mathbf{E}(X_1^k)$ term in a linear combination of the two previous lines and deduce an expression for $\mathbf{E}(X_2^k)$ only depending on $m_{Y,k}, m_{Z,k}$ and moments of $X_1$ and $X_2$ uniquelyof exponents less than $k$. Thereafter we can calculate $\mathbf{E}(X_1^k)$. This provides the moments of order $k$ and we can continue iteratively. q.e.d.
For necessity, two points with $v_1^2 = w_1^2$ is not enough: Thanks to Zeno44 for the good counterexample he provided; it extends to other cases with $v_1^2 = w_1^2$.
It is not clear to me what is the intuitive reason for the condition $v_1^2 \neq w_1^2$. The above "necessity" part is not the whole truth, since there is still the moment problem assumption. For this I can only provide the following comment.
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.