edit (2010-07-26): The question is much more involved than I'd originally thought. As implied in the question, I knew that congruence and similarity transformations are constructible. Immediately below this section is my original answer, which only demonstrates congruence transformations and was intended more to give an idea of what an answer might look like (since, at the time, there was another answer that was not particularly helpful). In the last section of this answer is my justification that all affine transformations of the plane are constructible. In re-reading that now, I realize that I'd assumed the ability to construct a point, say $P'$, on a line, say $\overleftrightarrow{RP}$, such that $\frac{RP'}{RP}$ is equal to some known ratio. This is equivalent to being able to construct the dilation of $P$ by the known ratio about center $R$. I've added the construction of such a dilation below the congruence transformation section.
edit (2012-01-28): A conversation with some colleagues reminded me about this problem and in starting to ask them about it, I realized I'd completely missed that all Möbius transformations are constructible. Since any Möbius transformation can be expressed as a composition of translation, reflection, inversion, dilation, rotation, and translation (I think there's a typical decomposition that's roughly in that order, hence my listing translation twice). The only one of these that I have not yet shown is constructible is inversion, so I have appended that construction.
As a partial answer, here are constructions of the basic congruence transformations, assuming basic construction techniques like constructing a line parallel or perpendicular to a given line through a given point and angle-copying:
reflection Given a point $P$ and a line $\ell$, construct the line perpendicular to $\ell$ through $P$, and construct the circle centered at the intersection of this new line and $\ell$ and passing through $P$. The image of $P$ under a reflection over the line $\ell$ is the point of intersection of the circle and the new line (the one not at $P$).
translation Given a point $P$ and a vector $\overrightarrow{AB}$ (from $A$ to $B$), construct the line through $A$ and $P$, the line through $B$ parallel to line $\overleftrightarrow{PA}$, and the line through $P$ parallel to $\overrightarrow{AB}$. The image of $P$ under translation by vector $\overrightarrow{AB}$ is the intersection of the two constructed parallels.
rotation Given a point $P$, a center of rotation $R$, and an $\angle ABC$ (from $A$ to $C$), construct the line through $P$ and $R$, copy $\angle ABC$ such that the copy of $B$ coincides with $R$ and the copy of $A$ is on ray $\overrightarrow{RP}$ and let the copy of $C$ be called $D$, construct the circle with center at $R$ and passing through $P$. The image of $P$ under rotation by $\angle ABC$ about point $R$ is the intersection of the circle with ray $\overrightarrow{RD}$.
dilation Given a point $P$, a center of dilation $R$, and a ratio $\frac{AC}{AB}$ (where point $C$ lies on ray $\overrightarrow{AB}$), translate $B$ to $B'$ and $C$ to $C'$ by the translation that takes $A$ to $R$, construct line $\overleftrightarrow{B'P}$, construct the line through $C'$ parallel to $\overleftrightarrow{B'P}$, and construct ray $\overrightarrow{RP}$. The image of $P$ under a dilation about $R$ by a factor of $\frac{AC}{AB}$ is the intersection of ray $\overrightarrow{RP}$ and the line through $C'$ parallel to $\overleftrightarrow{B'P}$.
All affine transformations are constructible. Per MathWorld and Wikipedia, an affine transformation of the plane is a transformation of the plane that preserves collinearity and preserves ratios of distances on any given line.
First, to show that affine transformations preserve parallelism, suppose that two lines $\overleftrightarrow{MN}$ and $\overleftrightarrow{PQ}$ are parallel, and that their images, lines $\overleftrightarrow{M'N'}$ and $\overleftrightarrow{P'Q'}$, intersect at $X'$, the image of $X$. Since affine transformations preserve collinearity, $X$ must be on $\overleftrightarrow{MN}$ and on $\overleftrightarrow{PQ}$, which is a contradiction, so $\overleftrightarrow{M'N'}$ and $\overleftrightarrow{P'Q'}$ cannot intersect. Thus, affine transformations preserve parallelism.
An affine transformation is determined by a $\triangle ABC$ and its image, $\triangle A'B'C'$ (per MathWorld; Wikipedia talks about defining an affine transformation by a parallelogram and its image, which is equivalent since affine transformations preserve parallelism). Given point $P$ and triangles $\triangle ABC$ and $\triangle A'B'C'$, construct line $\ell_1$ through $P$ parallel to $\overline{AB}$ and line $\ell_2$ through $P$ and parallel to $\overline{AC}$, call the intersection of $\ell_1$ with $\overline{AC}$ $I_1$ and call the intersection of $\ell_2$ with $\overline{AB}$ $I_2$, extend $\overline{A'B'}$ past $B'$ to a point $I'_2$ such that $\frac{AB}{AI_2}=\frac{A'B'}{A'I'_2}$, extend $\overline{A'C'}$ past $C'$ to a point $I'_1$ such that $\frac{AC}{AI_1}=\frac{A'C'}{A'I'_1}$, construct line $\ell'_1$ through $I'_1$ parallel to $\overline{A'B'}$ and line $\ell'_2$ through $I'_2$ parallel to $\overline{A'C'}$. The image of $P$ under the affine transformation mapping $\triangle ABC$ onto $\triangle A'B'C'$ is the intersection of lines $\ell'_1$ and $\ell'_2$.
inversion Given a point $P$ and a circle centered at $O$, construct ray $\overrightarrow{OP}$, let $X$ be the point of intersection of $\overrightarrow{OP}$ with the circle. The image of $P$ under an inversion through the circle is the image of $X$ under a dilation by $\frac{OX}{OP}$ centered at $O$.