= Solution
The displayed matrices describe the orientation-preserving component $SE(2)$ of the full <Euclidean group> $E(2)=O(2)\ltimes\mathbb R^2$. Both have the same <Lie algebra>, so this naming convention does not affect the calculation. Choose the translation generators $P_1,P_2$ and rotation generator $J$ in the <Lie algebra> of this <Matrix Lie group> as
$$
P_1=\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix},\qquad P_2=\begin{pmatrix}0&0&0\\0&0&1\\0&0&0\end{pmatrix},\qquad J=\begin{pmatrix}0&-1&0\\1&0&0\\0&0&0\end{pmatrix}.
$$
Taking matrix <commutators> gives
$$
\boxed{[P_1,P_2]=0,\qquad[J,P_1]=P_2,\qquad[J,P_2]=-P_1.}
$$
All remaining brackets follow from antisymmetry. The corresponding <left-invariant vector fields> in coordinates $(x,y,\psi)$ are
$$
L_1=\cos\psi\,\partial_x+\sin\psi\,\partial_y,\qquad L_2=-\sin\psi\,\partial_x+\cos\psi\,\partial_y,\qquad L_3=\partial_\psi.
$$
Their <Lie brackets of vector fields> obey the same relations. This also fixes which sign of the rotation generator is being used.
There is an important qualification to the printed connection prescription. For an arbitrary <vector field> $Z$ and smooth function $f$, its literal right-hand side obeys
$$
\lambda[L_i,fZ]=\lambda L_i(f)Z+\lambda f[L_i,Z].
$$
An <affine connection> instead requires $\nabla_{L_i}(fZ)=L_i(f)Z+f\nabla_{L_i}Z$. Thus \b[the formula for arbitrary $Z$ defines a connection only when $\lambda=1$]. For example, at $\psi=0$, take $i=1$, $f=x$ and $Z=L_1$: the erroneous formula gives $\lambda L_1$, whereas the <Leibniz rule> requires $L_1$.
The intended one-parameter family is the <bracket connection on a Lie group>: prescribe the rule for left-invariant $Z$, then extend it by the <Leibniz rule>. If $[L_i,L_j]=c_{ij}{}^kL_k$, this means
$$
\nabla_{L_i}L_j=\lambda c_{ij}{}^kL_k,\qquad \nabla_{L_i}(z^jL_j)=L_i(z^j)L_j+\lambda z^j c_{ij}{}^kL_k.
$$
Extend linearly over functions in the first slot. This is a well-defined <affine connection> for every real $\lambda$. The following <Ricci tensor> calculation applies to this intended family; under the literal arbitrary-$Z$ reading only its $\lambda=1$ member exists.
Use the curvature convention
$$
R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z,\qquad \operatorname{Ric}(X,Y)=\operatorname{tr}\bigl(Z\mapsto R(Z,X)Y\bigr).
$$
For <left-invariant vector fields> the <structure constants of a Lie algebra> are constant. The <Jacobi identity> gives
$$
\begin{aligned}
R(X,Y)Z&=\lambda^2\bigl([X,[Y,Z]]-[Y,[X,Z]]\bigr)-\lambda[[X,Y],Z]\\
&=\lambda(\lambda-1)[[X,Y],Z].
\end{aligned}
$$
Writing $A=\lambda(\lambda-1)$, the only nonzero curvature actions, apart from antisymmetry in the first two slots, are
$$
R(L_3,L_1)L_3=A L_1,\qquad R(L_3,L_2)L_3=A L_2.
$$
For instance, $[[J,P_1],J]=[P_2,J]=P_1$. Contracting the first and output slots gives $\operatorname{Ric}(L_3,L_3)=-2A$, while every other component vanishes. Equivalently, $[[Z,X],Y]=\operatorname{ad}_Y\operatorname{ad}_X Z$, so
$$
\operatorname{Ric}(X,Y)=A\,B(X,Y),\qquad B(X,Y)=\operatorname{tr}(\operatorname{ad}_X\operatorname{ad}_Y).
$$
This is the <Killing form>. Here $\operatorname{ad}_J$ rotates the two-dimensional translation space and vanishes on $J$, so $B(J,J)=-2$; translation generators give all other components zero. Therefore
$$
\boxed{(\operatorname{Ric}_{ij})_{i,j=1}^3=\begin{pmatrix}0&0&0\\0&0&0\\0&0&2\lambda(1-\lambda)\end{pmatrix}.}
$$
Since the coframe component dual to $L_3$ is $d\psi$, the same <Ricci tensor> is $2\lambda(1-\lambda)d\psi\otimes d\psi$ in these coordinates. Reversing the curvature convention reverses this sign.
At both $\lambda=0$ and $\lambda=1$, the full curvature vanishes. At $\lambda=0$, the <left-invariant vector fields> form a global <parallel frame>. At $\lambda=1$, every <right-invariant vector field> is parallel because left- and right-invariant <vector fields> commute; the connection is flat in that global <parallel frame>. Flatness does not imply zero torsion: for left-invariant arguments the <torsion tensor> is
$$
T(X,Y)=\nabla_XY-\nabla_YX-[X,Y]=(2\lambda-1)[X,Y].
$$
Thus \b[the two endpoint connections are flat, with opposite nonzero torsion]. For comparison, $\lambda=1/2$ is torsion-free but has nonzero curvature, with $\operatorname{Ric}_{33}=1/2$. No metric has been specified, so these connections should not automatically be identified with a <Levi-Civita connection>.
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