Let us think about the discrete case; that is let us suppose we are interested in determining a probability distribution $P$ on the discrete set
$\Omega ={\mathbb Z}_n^2$.

Such a probability distribution assigns a nonnegative weight to each $(i,j) \in \Omega$. $|\Omega| = n^2$, thus $P$ is determined by $n^2-1$ nonzero variables $p_{i,j}$ whose sum is less than $1$. To fix the marginals of $P$ means to put $2(n-1)$ constraints on $\{p_{i,j}\}$. In addition to these constraints, the present question also imposes a distribution on $X-Y$. These translate into $n-1$ further constraints on $p_{i,j}$. Thus, in general, $P$ will be a function of $n^2-1-3(n-1)$ free variables.

The special cases you mention (i.e, the clayton and gumbel copulas as well as the normal distribution), however, are determined by the marginal distributions and an additional real parameter. In general, if the given data makes sense, $\theta$ can be recovered by first writing an equation that it satisfies and then solving it.

Under any of the above mentioned copulas
the joint distribution equals $\Phi(F(x),G(y),\theta)$ where $F$ is the $X$ marginal, $G$ is the $Y$ marginal and $\theta$ is a real number. The only unknown here is $\theta$. Knowing the distribution of $X-Y$, means in particular we know the probability that $X-Y=n-1$[again assuming that we are operating in the discrete setup]. There is only one way this can happen, i.e, if $Y=0$ and $X=n-1$. Thus, we know the weight $p_{(n-1,0)}$ of the point
$(n-1,0)$. Then, $\theta$ is the solution of
$$\Phi(F(n-1),G(0),\theta) -\Phi(F(n-2),G(0),\theta) = p_{(n-1,0)}.$$

In the case of ${\mathbb R}^2$, one can for example, write the following equation for $\theta$:
$$
\int_{\mathbb R}^2 (x-y) d\Phi(F(x),G(y),\theta) = \int z dS(z)
$$
where $S$ is the distribution of $X-Y$ given in the problem.

Edit: more importantly, it seems, one has to check if the given distribution of $X-Y$ is compatible with the copula in question. Because, there are many equations that one can write for $\theta$ and all must give the same answer for the solution to be meaningful.