Let $\mu$ and $\nu$ be two finite measures on $\mathbb{R} ^d$. For every $s \in \mathbb{R} ^d$, $|s|=1$, denote by $\mu _s$ and $\nu _s$ the orthogonal projections of $\mu$ and $\nu$ respectively on the line $\{ as: a \in \mathbb{R} \}$. That is, for $a < b$ $$ \mu _s ( [a,b]) = \mu (\{x\in \mathbb{R} ^d: a \leq \langle x, s \rangle \leq b \}), $$ where $\langle , \rangle$ is the scalar product in $\mathbb{R} ^d$, and similar for $\nu _s$.
Question. Assume that $\mu _s = \nu _s$ for every $s \in \mathbb{R} ^d$, $|s|=1$. Does it follows that $\mu = \nu$?
The answer is obviously negative if $\mu$ and $\nu$ are not assumed to be finite (consider $\lambda$ and $2 \lambda$ with $\lambda$ being the Lebesgue measure).
Yes.
Let $s\in\mathbb R^d$. Then $$\mu(\mathbb R^d)=\mu_s(\mathbb R)=\nu_s(\mathbb R)=\mu(\mathbb R^d).$$
So they have the same mass. Of course if that common mass is $0$ then they are equal. Otherwise, up to dividing by their common mass we can suppose without loss of generality that they have mass 1.
So $\mu$ and $\nu$ are probability measures. Let $X$ and $Y$ be random variables whose probability distributions are $\mu$ and $\nu$ respectively. Then $\mu_s$, resp. $\nu_s$, is the distribution of $\langle X,s\rangle$, resp. $\langle Y,s\rangle$. Then by assumption, $\langle X,s\rangle$ and $\langle Y,s\rangle$ have the same distribution. Therefore $$ \mathbb E[\exp(i\langle X,s\rangle)]=\mathbb E[\exp(i\langle Y,s\rangle)]. $$ We deduce that $X$ and $Y$ have the same characteristic function. Therefore their distribution are equal, that is $\mu=\nu$.