$\newcommand\th\theta\newcommand\vpi\varphi\newcommand\R{\mathbb R}\newcommand\S{\mathbb S^{d-1}}$**Condensed version of the answer:** If $\th\cdot X\sim\mu_\th$ for all $\th\in\S$, then for each nonzero $t\in\R^d$ the distribution, say $\mu_t$, of $t\cdot X$ is determined by a simple rescaling of $\mu_\th$ (the case $t=0$ is trivial). On the other hand, by Bochner's theorem, a family $(\mu_t)_{t\in\R^d}$ of probability measures on $\R$ is the family of the distributions of the random variables $t\cdot X$ (for some random vector $X$ in $\R^d$) iff the function $\R^d\ni t\mapsto\vpi(t):=\int_{\R}e^{ix}\mu_t(dx)$ is positive definite and continuous. Thus, for any given family $(\mu_\th)_{\th\in\S}$ of probability measures on $\R$, your desired condition in question will hold if and only if $\vpi$ is positive definite and continuous.

**Detailed version of the answer:**
Suppose that
$$\text{there is a probability measure $\mu$ on $\R^d$ such that},\\
\text{if $X\sim\mu$, then $\th\cdot X\sim\mu_\th$ for all $\th\in\S$.}\tag{0}$$
Then
$$E_\mu f(\th\cdot X)=\int_{\R}f\,d\mu_\th \tag{1}$$
for all bounded continuous functions $f\colon\R\to\R$, where $E_\mu$ denotes the expectation assuming that $X\sim\mu$. In particular, (1) implies
$$E_\mu e^{ir\th\cdot X}=\int_{\R}e^{irx}\,\mu_\th(dx) \tag{2}$$
for all $r\in\R$ and all $\th\in\S$, so that
$$E_\mu e^{it\cdot X}=\vpi(t)$$
for all $t\in\R^d$,
where
$$\vpi(t):=
\left\{
\begin{aligned}
\int_{\R}e^{i|t|x}\,\mu_{t/|t|}(dx) &\text{ if }t\in\R^d\setminus\{0\}, \\
1 &\text{ if }t=0.
\end{aligned}
\right.
\tag{3}$$
So, if the condition (0) holds, then the function $\vpi$ -- being the characteristic function (c.f.) of $X$ -- is positive definite and continuous.

Vice versa, by Bochner's theorem, if $\vpi$ is positive definite and continuous, then there is a probability measure $\mu$ on $\R^d$ such that $E_\mu e^{it\cdot X}=\vpi(t)$ for all $t\in\R^d$, so that (2) holds for all $r\in\R$ and all $\th\in\S$, that is, the c.f. of $\th\cdot X$ is the same as the c.f. of $\mu_\th$, which means that $\th\cdot X\sim\mu_\th$. So, (0) holds if the function $\vpi$, defined by (3), is positive definite and continuous.

Thus, for any given family $(\mu_\th)_{\th\in\S}$ of probability measures on $\R$, condition (0) will hold if and only if $\vpi$ is positive definite and continuous.

Somewhat related to this is the work by Shepp and his co-authors on probabilistic tomography, about partial restoration of the distribution of a random vector $X$ in $\R^d$ based on the known distributions of a finite number of linear functionals of $X$; see e.g. this paper and references there.