Real-valued 2D Fourier series?

Yes! And these types of expansions occur in a variety of applications, e.g., solving the heat or wave equation on a rectangle with prescribed boundary and initial data.

As a specific example, we can think of the following expansion as a two dimensional Fourier sine series for $f(x,y)$ on $0<x<a$, $0<y<b$: $$ f(x,y)=\sum_{n=1}^\infty \sum_{m=1}^\infty c_{nm}\sin\left({n\pi\, x\over a}\right)\sin\left({m\pi\, y\over b}\right), \quad 0<x<a,\ 0<y<b, $$ where the coefficients (obtained from the same type of orthogonality argument as in the 1D case) are given by \begin{align} c_{nm}&={\int_0^b \int_0^a f(x,y)\sin\left({n\pi\, x\over a}\right)\sin\left({m\pi\, y\over b}\right)\,dx\,dy\over \int_0^b \int_0^a \sin^2\left({n\pi\, x\over a}\right)\sin^2\left({m\pi\, y\over b}\right)\,dx\,dy}\\ &={4\over a b}\int_0^b \int_0^a f(x,y)\sin\left({n\pi\, x\over a}\right)\sin\left({m\pi\, y\over b}\right)\,dx\,dy, \quad n,m=1,2,3,\dots \end{align}

For example, the picture below shows (left) the surface $$f(x,y)=30x y^2 (1-x)(1-y)\cos(10x)\cos(10y), \quad 0<x<1,\ 0<y<1,$$ and a plot of the two dimensional Fourier sine series (right) of $f(x,y)$ for $n,m,=1,\dots,5$:

Finally, keep in mind that we are not limited just to double sums of the form sine-sine. We could have any combination we like so long as they form a complete orthogonal family on the domain under discussion.

The full real-valued 2D Fourier series is: $$ \begin{align} f(x, y) & = \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\alpha_{n,m}cos\left(\frac{2\pi n x}{\lambda_x}\right)cos\left(\frac{2\pi m y}{\lambda_y}\right) \\ & + \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\beta_{n,m}cos\left(\frac{2\pi n x}{\lambda_x}\right)sin\left(\frac{2\pi m y}{\lambda_y}\right) \\ & + \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\gamma_{n,m}sin\left(\frac{2\pi n x}{\lambda_x}\right)cos\left(\frac{2\pi m y}{\lambda_y}\right) \\ & + \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\delta_{n,m}sin\left(\frac{2\pi n x}{\lambda_x}\right)sin\left(\frac{2\pi m y}{\lambda_y}\right) \\ \end{align} $$

The coefficients are found with: $$ \alpha_{n,m} = \frac{\kappa}{\lambda_x \lambda_y}\int_{y_0}^{y_0+\lambda_y}\int_{x_0}^{x_0+\lambda_x}f(x,y)cos\left(\frac{2\pi n x}{\lambda_x}\right)cos\left(\frac{2\pi m y}{\lambda_y}\right)dx dy \\ \beta_{n,m} = \frac{\kappa}{\lambda_x \lambda_y}\int_{y_0}^{y_0+\lambda_y}\int_{x_0}^{x_0+\lambda_x}f(x,y)cos\left(\frac{2\pi n x}{\lambda_x}\right)sin\left(\frac{2\pi m y}{\lambda_y}\right)dx dy \\ \gamma_{n,m} = \frac{\kappa}{\lambda_x \lambda_y}\int_{y_0}^{y_0+\lambda_y}\int_{x_0}^{x_0+\lambda_x}f(x,y)sin\left(\frac{2\pi n x}{\lambda_x}\right)cos\left(\frac{2\pi m y}{\lambda_y}\right)dx dy \\ \delta_{n,m} = \frac{\kappa}{\lambda_x \lambda_y}\int_{y_0}^{y_0+\lambda_y}\int_{x_0}^{x_0+\lambda_x}f(x,y)sin\left(\frac{2\pi n x}{\lambda_x}\right)sin\left(\frac{2\pi m y}{\lambda_y}\right)dx dy $$ $$ \begin{align} \text{Where } \kappa & = 1 \text{ if } n = 0 \text{ and } m = 0 \\ & = 2 \text{ if } n = 0 \text{ or } m = 0\\ & = 4 \text{ if } n> 0 \text{ and } m > 0 \end{align} $$ Example plot

@JohnD only details about the coefficients. The correct formula is:

$$c_{n,m} = \frac{\int_{0}^a \int_0^b f(x,y)sin(\frac{n\pi x}{a})sin(\frac{n\pi y}{b})dxdy}{{\int_{0}^a \int_0^b sin^2(\frac{n\pi x}{a})sin^2(\frac{n\pi y}{b})dxdy}{}}$$

the impression that $c_{n,m}$ is $1$. That's not true. Cheers!