Let $u(x,y)=X(x)Y(y)$, then plug that in to your original equations
$$
X'(x)Y(y) + X(x)Y'(y) = 0
$$
Dividing through by $X(x)Y(y)$ and rearranging
$$
\frac{X'(x)}{X(x)} = -\frac{Y'(y)}{Y(y)}
$$
The left hand side depends only on the variable $x$, and the right hand side depends only on the variable $y$, yet they are equal to each other for all $(x,y)$. Each side must therefore be a constant (call it $\lambda$), giving two equations
$$
\frac{X'(x)}{X(x)} = \lambda
\\
\frac{Y'(y)}{Y(y)} = -\lambda
$$
Rearranging gives
$$
X'(x) - \lambda X(x) = 0
\\
Y'(y) + \lambda Y(y) = 0
$$
These are first order ODEs you can solve. The solutions are
$$
X(x) = e^{\lambda x}
\\
Y(y) = e^{-\lambda y}
$$
A general solution (for any $\lambda$) is
$$
u_\lambda(x,y) = c(\lambda)e^{\lambda x}e^{-\lambda y} = c(\lambda)e^{\lambda (x-y)}
$$
Without boundary conditions, this is as far as you can go. The boundary conditions would help you pin down what $\lambda$ and $c(\lambda)$ are. For an equation with at least two derivatives, that's also when the Fourier series side of it comes into play. Usually you get a discrete spectrum of $\lambda$ values, and the solution is a summation of terms like that over the different values of $\lambda$. Problems on unbounded domains can have a continuous spectrum of $\lambda$ values and you end up with an integral instead of a summation (a Fourier transform instead of a Fourier series essentially).
For this problem, with the right interpretation of the terms, the general solution will always be
$$
u(x,y) = \int_{-\infty}^\infty c(\lambda)e^{\lambda (x-y)}\;d\lambda
$$
which can reduce to a discrete summation with a bit of help from complex analysis.
