I conjecture that every shear transformation of the plane is geometrically congruent to an orthogonal dilation of the plane. That is, in Euclidean geometry of the plane, if I shear figure $F$, I can produce a geometrically congruent figure $F'$ via orthogonal dilation.
Is my conjecture correct? If so, what is the orthogonal dilation $D$ congruent to $$\begin{bmatrix}1 & \tan \theta \\ 0 & 1\end{bmatrix}.$$ Since shears have determinant 1, I believe it must be of the form $$\begin{bmatrix}\alpha & 0 \\ 0 & \frac 1 \alpha \end{bmatrix},$$ but can't figure out $\alpha$ as a function of $\theta$ (or of $\lambda$ in the general case above).
Likewise, consider 
– what dilation, along any axes, can produce it? I haven't been able to sketch one.
The support for this conjecture is via linear algebra; below are the definitions and support:
Definitions
Shear transformation means $S: \mathbb R^2 \to \mathbb R^2, S = \begin{bmatrix}1 & \lambda \\ 0 & 1\end{bmatrix}$ or $S = \begin{bmatrix}1 & 0 \\ \lambda & 1\end{bmatrix}$.
Congruent is defined: a set of points $F$ is congruent to $F'$ if there is an isometry (preserving distance, angles, and colinearity) that sends $F \to F'$. Note that congruent does not imply matrix similarity: it a geometric property related by isometry.
Orthogonal dilation means an arbitrary set of axes is chosen, each orthogonal to the rest, and the plane is scaled by a (perhaps different) constant along each. This is precisely the matrices of the form $PDP^{-1}$ where $P$ is orthogonal and $D$ is diagonal.
Conjecture
I conjecture that for any shear $S$ there exists an orthogonal dilation $D$ such that for all figures $F$, $S(F) \cong D(F)$.
Support
Every real 2×2 matrix can be decomposed into a polar decomposition $A = PU$ where $U$ is orthogonal diagonizable and $P$ is an orthogonal matrix (and therefore an isometry).
This implies the conjecture and also something related: Every dilation along non-orthogonal axes is congruent to a dilation along congruent axes.
Background: While trying to determine Geometrically, what is unique about the mapping of an orthogonally diagonizable matrix (versus non-orthogonal)? , I came to the conclusion that there is no difference: every diagonalizable transform is geometrically congruent to an orthogonally diagonalizable geometric transform!
Further background to this conjecture is at Is every symmetric matrix in $\mathbb R^n$ a (non-uniform) stretch? .
Best Answer
What you call an "orthogonal dilation" is what is normally called an "orthogonally diagonalizable transformation" (or matrix).
TL;DR: the numbers $\alpha$ and $\frac{1}{\alpha}$ are the singular values of $A$.
As you note, the polar decomposition of a matrix says:
Theorem. For any real square matrix $A$ there exists an orthogonal matrix $W$ and a positive semidefinite matrix $P$ such that $A=WP$. Moreover, if $A$ is invertible then the representation is unique.
A positive semidefinite matrix is a matrix of the form $B^TB$ for some square matrix $B$; in particular, positive semidefinite matrices are always symmetric, and therefore orthogonally diagonalizable.
This says that every real square matrix can be decomposed as an orthogonally diagonalizable matrix followed by an orthogonal matrix (which is necessarily a rigid motion); this rigid motion realizes your "congruence".
So you already know that your "conjecture" holds; it seems that you are just trying to figure out what the diagonal matrix (or perhaps what the positive definite matrix $P$?) is.
If we let $A$ be a matrix of the form $$A = \left(\begin{array}{cc} 1 & \lambda\\ 0 & 1\end{array}\right),$$ then $A=WP$ will mean that $A$ corresponds to an orthogonally diagonalizable matrix (with positive eigenvalues, in fact), followed by an orthogonal linear transformation (which is necessarily a rigid motion).
Since your $A$ is invertible, the representation is unique, and we can obtain an explicit form for $P$ using the singular value decomposition of $A$. We can also calculate the value of $\alpha$ (in your notation at the top) in terms of $\lambda$ straightforwardly enough.
Namely, if $A=U\Sigma V^T$ is a singular value decomposition for $A$, then $W=UV^T$ and $P=V\Sigma V^T$ will be the desired expression. Here $\Sigma$ is a diagonal matrix whose diagonal entries are the singular values of $A$.
The singular values of $A$ are the square roots of the eigenvalues of $A^TA$. We have $$A^TA = \left(\begin{array}{cc} 1 & 0\\ \lambda & 1\end{array}\right) \left(\begin{array}{cc} 1 & \lambda\\ 0 & 1\end{array}\right) = \left(\begin{array}{cc} 1 & \lambda\\ \lambda & 1+\lambda^2 \end{array}\right).$$ The eigenvalues are then the roots of $t^2 - (\lambda^2+2)t+1$, which are $$\frac{\lambda^2 + 2 \pm \lambda\sqrt{\lambda^2+4}}{2}.$$ So the matrix $\Sigma$ will be $\Sigma=\mathrm{Diag}(\alpha_1,\alpha_2)$, where $$\begin{align*} \alpha_1 &= \sqrt{\frac{\lambda^2+2 + |\lambda|\sqrt{\lambda^2+4}}{2}},\\ \alpha_2 &= \sqrt{\frac{\lambda^2+2-|\lambda|\sqrt{\lambda^2+4}}{2}} \end{align*}$$ (following the convention that the diagonal entries of $\Sigma$ must be non-increasing in size). It is straighforward to verify that $\alpha_1\alpha_2=1$.
Symmetrically, if $A=\left(\begin{array}{cc}1&0\\ \lambda&1\end{array}\right)$, then $$A^TA=\left(\begin{array}{cc} 1+\lambda^2 & \lambda\\ \lambda & 1\end{array}\right).$$ This has the same characteristic polynomial, hence the same eigenvalues, hence the same singular values, you again get $\alpha_1$ and $\alpha_2$.