Classifying the solutions is a non-trivial problem when two of the $\lambda_i$ are equal, even when they are constants. The reason is that this is essentially an overdetermined problem when two of the $\lambda_i$ are equal.

The point is this: When the $\lambda_i$ are distinct, a choice of an orthonormal coframing $\omega = (\omega_i)$ is essentially a map of $\mathbb{R}^3$ into $\mathrm{SO}(3)$, i.e., a choice of three functions of 3 variables. The requirement that the metric $g$ as defined above be flat is three equations, i.e., the vanishing of the three symmetric functions of the eigenvalues of the Ricci tensor of $g$. It's not surprising that this system of equations has local solutions (when the $\lambda_i$ are distinct and real-analytic). Setting it up as an exterior differential system, one quickly finds that this system is involutive, and the Cartan-Kähler theorem does the job. (N.B.: This fact is not special to $3$-dimensions or to flatness. The Cartan-Kähler analysis I describe in this MO answer generalizes without significant change to cover the case of not-necessarily-constant prescribed real-analytic singular value functions $\lambda_1<\lambda_2<\cdots<\lambda_n$ on $M$ for real-analytic local mappings $f:(M^n,g)\to (N^n, h)$ between any two real-analytic Riemannian manifolds.)

However, when $\lambda_1=\lambda_2\not=\lambda_3$, the situation is entirely different. Now, one doesn't need to specify a complete orthonormal coframe, just $\omega_3$, a unit $1$-form, which is essentially a map of $\mathbb{R}^3$ into $S^2$, i.e., two functions of $3$ variables. Now, if $g_0$ is the Euclidean metric, we can write $g$ in the form
$$
g = e^{\lambda_1} g_0 + (e^{\lambda_3}-e^{\lambda_1})\,{\omega_3}^2,
$$
and the flatness condition on $g$, i.e., $\mathrm{Ric}(g) = 0$ is still three equations on the two unknowns that go into the definition of $\omega_3$. For generic $\lambda_1\not=\lambda_3$, one does not expect there to be *any* solutions, even locally, but constructing an actual example of a pair $(\lambda_1,\lambda_3)$ for which no solution exists might be hard. (As a comparison, consider the even more restrictive case in which all the $\lambda_i$ are equal. There, one definitely gets conditions on the function $\lambda = \lambda_1=\lambda_2=\lambda_3$ in order for $e^\lambda g_0$ to be flat. In fact, there is only a 4-parameter family of (local) functions $\lambda$ that satisfy these conditons, and, of course, we know these explicitly.) I expect that there will be *some* set of PDE relating $\lambda_1$ and $\lambda_3$ whose satisfaction tells whether solutions exist, but I have a feeling that this compatibility condition is liable to be quite complicated.

Even when $\lambda_1$ and $\lambda_3$ are constants, classifying the solutions is not trivial. Clearly, one can reduce to the case in which $\lambda_1=0$ by scaling $g$ by a constant (which won't affect its flatness). For simplicity, set $e^{\lambda_3} = a^2\not=1$. As Willie Wong has observed, taking $\omega_3$ to have constant coefficients always gives a solution, which one can regard as 'trivial'.

Meanwhile, there are nontrivial solutions: Consider the following construction: Let $I\subset\mathbb{R}$ be an interval and let $f = (f_i):I\to S^2$ be a smooth mapping. Consider, the $1$-parameter family of metrics
$$h_{f,b}= du^2 + dv^2 + b^2\,\bigl(f_1(t)\,u + f_2(t)\,v + f_3(t)\bigr)^2\,dt^2$$
in the domain $D_f\subset\mathbb{R}^2\times I$ where $h_{f,b}$ is positive definite, i.e., the complement of the locus where $f_1(t)\,u + f_2(t)\,v + f_3(t)=0$. Here $b>0$ is a constant parameter. It is simple to verify that $h_{f,b}$ is a flat metric for any choice of $f$ and $b$. As a result, there is a smooth local isometry $\phi_f:D_f\to\mathbb{R}^3$ so that $\phi_f^*(g_0) = h_{f,1}$, where $g_0$ is the standard metric on $\mathbb{R}^3$. Restricting the domain of $\phi_f$ to some open subset $D'_f\subset D_f$, if necessary, one can arrange that $\phi_f:D'_f\to\mathbb{R}^3$ be injective. Hence there will exist a unique quadratic form
$g_{\lambda_3}$ on the (open) image $\phi_f(D'_f)\subset\mathbb{R}^3$ such that $\phi_f^*(g_{\lambda_3}) = h_{f,a}$. By construction,
$$
g_{\lambda_3} = g_0 + (e^{\lambda_3}-1)\,{\omega_3}^2,
$$
where $\omega_3$ is a $1$-form (of unit size with respect to $g_0$); in fact,
$$
\phi_f^*(\omega_3) = \bigl(f_1(t)\,u + f_2(t)\,v + f_3(t)\bigr)\,dt.
$$
It's easy to show that this $g_{\lambda_3}$ is a 'trivial' solution if and only if $f_1 = f_2 = 0$, i.e., $f:I\to S^2$ is the constant map to either the north or south pole.

Thus, this construction gives a class of local solutions that essentially depend on two functions of one variable.

The remaining question is whether there are any solutions that are *not* equivalent to one of these up to isometries in $\mathbb{R}^3$.

What one can say is that, if there are any such solutions for given constants $\lambda_1 = 0$ and $\lambda_3\not=0$, there are not very many, even locally.
More precisely, using the Cartan structure equations, one can show the following: Suppose that one has a unit $1$-form $\omega_3$, such that
$$
g = g_0 + (e^{\lambda_3}-1)\,{\omega_3}^2
$$
is flat (where $\lambda_3$ is a nonzero constant). Let $X$ be the vector field $g_0$-dual to $\omega_3$ and consider the quadratic form
$$
Q = \mathscr{L}_X\bigl(g_0-{\omega_3}^2\bigr).
$$
If $Q$ is identically zero, then $g$ arises locally from the above construction using an $f:I\to S^2$. Moreover, there is at most a $6$-dimensional space of local solutions $\omega_3$ that have a non-zero $Q$. (Probably, the space of such solutions has considerably smaller dimension, but I don't know how much smaller.)