TikZ: How to draw an isometric drawing in tikz

Like Andrew wrote, 3d is not in the documentation. I updated my answer because I introduced some mistakes and complications. First we need to define the vectors for the xyz system, then with the 3dlibrary options, we can work in a specific plane.

 canvas is xy plane at z=0

I design the plane xy with z=0, I made a mistake with yx because in this case I exchange the vectors x and y.

Here a list of the options in 3d

coordinate system  xyz cylindrical
coordinate system  xyz spherical

/tikz/cs/longitude/
/tikz/cs/latitude/

 plane origin
 plane x
 plane y

 canvas is plane
 canvas is xy plane at z
 canvas is yx plane at z
 canvas is xz plane at y
 canvas is zx plane at y
 canvas is yz plane at x
 canvas is zy plane at x

The code :

 \documentclass{scrartcl}
 \usepackage{tikz}
 \usetikzlibrary{arrows,3d}

 % see the explanation below
 \makeatletter
 \tikzoption{canvas is xy plane at z}[]{%
   \def\tikz@plane@origin{\pgfpointxyz{0}{0}{#1}}%
   \def\tikz@plane@x{\pgfpointxyz{1}{0}{#1}}%
   \def\tikz@plane@y{\pgfpointxyz{0}{1}{#1}}%
   \tikz@canvas@is@plane
 }
 \makeatother 

\begin{document}
\begin{tikzpicture}
 [x={(0.866cm,0.5cm)}, y={(-0.866cm,0.5cm)}, z={(0cm,1cm)}, scale=2]
  \draw[->,red] (0,0,0) --  (1,0,0);
  \draw[->,red] (0,0,0) --  (0,1,0);    
  \draw[->,red] (0,0,0) --  (0,0,1);    
  \begin{scope}[canvas is xy plane at z=0]
      \draw[blue,shift={(1.5,0)}] (0,0) -- (1,0)--(1,1)--(0,1)--cycle;
      \draw[blue,shift={(3,0)}] (0,0) -- (1,0)--(1,1)--(0,1)--cycle;
  \end{scope}  
  \end{tikzpicture}

  \end{document}

The code for the option used in my example is

  \tikzoption{canvas is xy plane at z}{%
    \tikz@addtransform{\pgftransformshift{\pgfpointxyz{0}{0}{#1}}}%
  } 

This is only a shift transformation.

Update

As Jake noticed in this answer grid in 3d

The implementation of canvas is xy plane at z in tikzlibrary3d.code.tex is incorrect, it merely sets a coordinate shift, but doesn't activate the full transformation code necessary. You can redefine the key correctly within your document:

 \makeatletter
 \tikzoption{canvas is xy plane at z}[]{%
   \def\tikz@plane@origin{\pgfpointxyz{0}{0}{#1}}%
   \def\tikz@plane@x{\pgfpointxyz{1}{0}{#1}}%
   \def\tikz@plane@y{\pgfpointxyz{0}{1}{#1}}%
   \tikz@canvas@is@plane
 }
 \makeatother  

I forgot this and it is why I use yxinstead of xy in my first attempt.

As @Jake pointed out in a comment, you can specify the coordinate system of your choice as an option of the tikzpicture environment. Here is an example:

\documentclass[a4paper,11pt]{article}

\usepackage{tikz}

\begin{document}
\begin{tikzpicture}[y={(-1cm,0.5cm)},x={(1cm,0.5cm)}, z={(0cm,1cm)}]
% coordinate system
\coordinate (O) at (0, 0, 0);
\draw[-latex] (O) -- +(1, 0,  0) node [right] {$x$};
\draw[-latex] (O) -- +(0,  1, 0) node [left] {$y$};
\draw[-latex] (O) -- +(0,  0, 1) node [above] {$z$};
% rectangles
\draw (3,-1.5,0) -- (3,1.5,0) -- (5,1.5,0) -- (5,-1.5,0) -- cycle;
\draw (6,-1.5,0) -- (6,1.5,0) -- (8,1.5,0) -- (8,-1.5,0) -- cycle;

\end{tikzpicture}
\end{document}

Instead of giving the base vectors as a (x,y) tuple [y={(-1cm,0.5cm)},x={(1cm,0.5cm)}, z={(0cm,1cm)}], I prefer to use (angle:length)

\begin{tikzpicture}[x={(90:1cm)}, y={(0:1cm)}, z={(45:0.7cm)}]

This gives you a better understanding of the vector directions.