Do all projections matrices take this form?

$A(A^TA)^{-1}A^T$ is symmetric, but not all projection matrices are symmetric --- such as $\pmatrix{1&1\\ 0&0}$. Thus the answer is clearly no.

It is true, however, that all orthogonal (with respect to the usual inner product) projection matrices over $\mathbb R$ can be written in the form of $A(A^TA)^{-1}A^T$. By definition, if $P\in M_n(\mathbb R)$ is an orthogonal projection, then $P|_U=\operatorname{id}$ and $P|_{U^\perp}=0$ for some subspace $U\subseteq\mathbb R^n$. Let $A$ be any matrix whose columns form a basis of $U$ (any basis will do; it doesn't have to be orthonormal). Then $A^TA$ is nonsingular and $A(A^TA)^{-1}A^Tv=0$ for every $v\in U^\perp$. Also, since the columns of $A$ span $U$, every vector $u\in U$ can be written as $Ax$ for some $x\in\mathbb R^n$. Therefore $$ \left(A(A^TA)^{-1}A^T\right)u=\left(A(A^TA)^{-1}A^T\right)(Ax)=\left(A(A^TA)^{-1}A^TA\right)x=Ax=u $$ for every $u=Ax\in U$. Hence $P$ and $A(A^TA)^{-1}A^T$ agree everywhere on $\mathbb R^n$, i.e. $P=A(A^TA)^{-1}A^T$.

As pointed out by @user1551, this is only true for orthogonal projection matrices.

Let $P$ be the orthogonal projection operator that projects a vector $b \in \mathbb R^n$ onto a subspace $S \subset \mathbb R^n$. Let $(a_1,\ldots,a_m)$ be a basis for $S$, and let $A$ be the matrix whose $i$th column is $a_i$. Then $S =\{Ax \mid x \in \mathbb R^m\}$, and projecting $b$ onto $S$ is equivalent to selecting $x$ so as to minimize the distance from $b$ to $Ax$. Equivalently, we want to minimize $$ r(x) = \| Ax - b \|^2. $$ This is a least squares problem. Setting the gradient equal to $0$, we find that $x$ satisfies $$ \tag{1} A^T(Ax-b) = 0 $$ or equivalently $$A^TA x = A^T b.$$ (This system of equations is often called the "normal equations". Visually, equation (1) just says that the residual vector $b - Ax$ is orthogonal to the column space of $A$.)

It follows that $x = (A^T A)^{-1} A^T b$. So the projection of $b$ onto $S$ is $$ P(x) = Ax = A(A^T A)^{-1} A^T b. $$