We know two independent random variables $X,Y$ has a common distribution, and we know both have zero expectation (or some other convenient centering). We don't know what that common distribution is, but we do know the distribution of $D=X-Y$.
Is it possible to determine the distribution function of $X$ and $Y$?
Let us assume that $X,Y,D$ has continuous distributions, with cumulative distribution function (cdf) and density $F,f$ (for $X,Y$) and $G,g$ (for $D$). First, note that $D$ always have a symmetric distribution:
$$ \newcommand\myeq{\mathrel{\stackrel{{\mbox{ D}}}{=}}}
X-Y \myeq Y-X \myeq -(X-Y)
$$ since $X\myeq Y$, where the symbol $\myeq$ means "have the same distribution". So for the problem to be well posed, it is necessary that we postulate a symmetric distribution for the difference $D$.
Now
$$
G(u) =P(X-Y \le u) = \int_{-\infty}^\infty P(X \le u+Y \mid Y=y) f(y)\; dy \\
= \int F(u+y) f(y) \, dy
$$
and by differentiation wrt $u$ under the integral sign we get
$$
g(u) = \int f(u+y) f(y) \; dy
$$
which is an integral equation. But, for a probability problem it is easier to formulate this in terms of the moment generating function (mgf) of the difference, assuming that it exists. If it do not exist, we can work in like manner using the characteristic function. So assume the difference have mgf $M(t)$ and $X,Y$ have common mgf $G(t)$. Since $D$ has a symmetric distribution, we have
$$ \DeclareMathOperator{\E}{\mathbb{E}}
M(t) = \E e^{tD} = \E e^{-tD} = M(-t)
$$ so that necessarily $M(t)=M(-t)$. We also find
$$
M(T)= \E e^{t(X-Y)} = \E e^{tX} \E e^{-tY} = G(t) G(-t)
$$
giving the equation $M(t) = G(t) G(-t)$ which we can try to solve for $G(t)$. But this is not really enough, since nothing guarantees that a function $G(t)$ we find that way is a mgf for a probability distribution!
Let us look at some examples. If $D$ has a centered normal distribution with variance 2, its mgf is $M(t)= e^{t^2} = \exp\left( \frac12 t^2 + \frac12 t^2\right)$ so we find the solution $G(t) = e^{\frac12 t^2}$ which indeed is the mgf of the standard normal distribution.
If $D$ has a centered triangular distribution with density function
$$
f(x) =\begin{cases} 1-|x|,& |x| \le 1 \\
0, & |x| > 1
\end{cases}
$$
then its mgf is $M(t) = \frac{e^t-2+e^{-t}}{t^2}$ which can be factored as
$M(t) = \left(\frac{e^{t/2} - e^{-t/2}}{t}\right)^2$ giving the factor
$G(t) = \frac{e^{t/2} - e^{-t/2}}{t}$ which is indeed the mgf of the uniform distribution on the interval $(-1/2, 1/2)$.
Now let $D$ have the symmetric Laplace distribution with mgf (see wikipedia) $M(t)=\frac1{1-t^2}= \frac1{1-t} \cdot \frac1{1+t}=G(t)G(-t)$
with $G(t)=\frac1{1-t}$ which is indeed the mgf of an exponential distribution. In the first two examples, our solution was symmetric, but in this third example the solution is an asymmetric distribution. Is there in this case also a symmetric solution? we could try the alternative solution
$$
G(t) =\sqrt{\frac1{1-t^2}}
$$
But, I do not know if this is a valid mgf of some probability distribution.
If it is, we would have shown that this problem do not necessarily have a unique solution.
The OP did not tell us in which form he has information on the distribution of $D$. If he has an iid sample from $D$, maybe he could calculate an empirical estimate of the moment generating function, estimating $G(t)$ by taking the square root of the estimate of $M(t)$, and try to approxiamtely invert that by using a saddle point approximation, see How does saddlepoint approximation work?
If he wants a better answer we need to know in which form is his information.
