Self-similar solutions
Notice that the the solution is invariant under the one-parameter transformation $$\phi(x,t)\mapsto \mu \phi(\mu x, \mu t), \mu\in\mathbb R^+$$
which we can interpret as an action of a Lie group, as well as translations in time, which means we can put $\mu=\frac{1}{T-t}$ to obtain self-similar solutions $\phi(x,t)=\frac{1}{T-t} f(\xi)$, where $\xi:=\frac{x}{T-t}$. $T$ is some characteristic "blowup" time.
After a slightly laborious calculation, the original equation transforms into an ODE, $$(1+\xi^2)f''(\xi)-4\xi f'(\xi)-2f(\xi)+\lambda [f(\xi)]^3=0.$$
It's formidable, but at least admits a constant solution $f(\xi)=\sqrt{2/\lambda}$.
Other solutions grow in $\xi$. For instance, here is the plot and the phase portrait of the solution with data $(f(0)=1,f'(0)=0,\lambda=1)$, crammed up to conserve some space!
Note that here the you need to interpret $\xi$ carefully.
Moreover, the 2nd order ODE may be recast into a system of 1st order ODEs
$$\frac{d f}{d\xi}=v, \:\frac{d v}{d\xi}=\frac{f(2-\lambda f^2)+4\xi v}{1+\xi^2}$$
The system is nonautonomous but we may cheat by analysing its behaviour at a fixed $\xi$. The critical points where the LHS's vanish are $(f,v)=(0,0)$ and $(\sqrt{2/\lambda},0)$, for all $\xi$. The eigenvalues turn out to be positive and negative for all $\xi$, meaning it's a saddle point, and complex but with $\Re{\lambda_{\pm}}>0$, meaning it's again unstable. Therefore all such solutions blow up.
Travelling solutions
We set $\eta=x-ct$ and try out solutions of the form $\phi(x,t)=g(\eta)$. With this the equation is recast as $$g''(\eta)=\tilde\lambda [g(\eta)]^3$$
where $\tilde\lambda:=\lambda/(c^2-1)$. Similarly as with trig/hyperbolic functions, this either blows up or oscillates, depending on the sign of $\tilde\lambda$. For negative $\tilde\lambda$, a solution is given in terms of the Jacobi elliptic function
$$g(\eta)=A\,cn( A \sqrt{-\tilde\lambda}\eta,1/\sqrt{2})$$
This solution is known as a compact breather. It's periodic, but probably highly unstable. The velocity and amplitude are probably independent parameters.
So we can try the ansatz $\phi(x,t)=A(x,t)e^{i(x-ct)}$ which leads to, as you have observed, $$(1-c^2)A_x+A_{tt}-A_{xx}+2i(A_x-c A_t)=\lambda |A|^2 A,$$
which is more difficult than the original equation unless further assumptions are made. If we insist on $A(x,t)$ being real, this implies $A_x=c A_t$. We want to avoid having an amplitude that is also of the form of a travelling wave, since that leads to the equation in the previous paragraph. We can check this by trying an ansatz $\phi(\eta,t)=A(\eta)e^{i(n \eta -mt)}$, $\eta=x-ct$, giving us
$$c^2 A''(\eta)-2 i c (-c m-n) A'(\eta)-(-c m-n)^2 A(\eta)-A(\eta)=\lambda |A(\eta)|^2 A(\eta)$$
where we can choose $cm+n=0$ to eliminate the $A'$ term, implying $A(\eta)$ is real, and giving us
$$c^2 A''(\eta)=\lambda A(\eta)^3+A(\eta)$$
This can be integrated if we mulitply by $ 2/c^2 \cdot A'$ giving us
$$A'^2=(\alpha_0+\alpha_1 A^2)(\beta_0+\beta_1 A^2)\Rightarrow \eta=\int^{A(\eta)}_0 \frac{\mathrm d \tilde A}{\sqrt{(\alpha_0+\alpha_1 \tilde A^2)(\beta_0+\beta_1 \tilde A^2)}}$$
which is essentially the same as before, the solution being given in terms of a Jacobi elliptic function as $$A(\eta)=\sqrt{\frac{\alpha_0}{\alpha_1}} sn(\sqrt{\alpha_1\beta_0}\eta, (\alpha_0\beta_1)/(\alpha_1\beta_0))$$
Linearisation; Stable and self-modulating solutions
Let's again try the ansatz $\phi(x,t)=A(x,t)e^{i(\omega_0 t - k_0 x)}$. I will change the original equation slightly, into $\phi_{tt}-c^2\phi_{xx}=\lambda |\phi|^2\phi$, where $c=\omega/k$ is the natural phase velocity of the free linear wave:
$$A_{tt}-c^2 A_{xx}+2 i (c^2 k_0 A_x+\omega_0 A_t)+(c^2k_0^2-\omega_0^2)A-\lambda |A|^2 A=0$$
introducing group velocity $c_g=c^2\frac{k_0}{\omega_0}=\frac{d\omega}{dk}\big|_{k=k_0}$ and a coupling parameter $\alpha=\lambda/(2\omega_0)$
$$\frac{1}{2\omega_0}(A_{tt}-c^2 A_{xx})+ i(c_g A_x+ A_t)+\frac{\omega_0}{2}\left((\frac{c_g}{c})^2-1\right)A-\alpha |A|^2 A=0$$
which we can further simplify by putting $\rho=\frac{\omega_0}{2}\left((\frac{c_g}{c})^2-1\right)$.
Now we switch to coordinates moving at group velocity, $\xi=x-c_g t,\tau=t$, so that we are essentially looking at how the envelope behaves:
$$i A_{\tau}-\beta A_{\xi\xi}+\rho A-\alpha |A|^2 A = 0$$
where $\beta=c_g^2-c^2=\frac{2 c^2}{\omega_0}\rho= \frac{1}{2}\frac{d^2\omega}{dk^2}\big|_{k=k_0}$ is a dispersive parameter. This is a nonlinear Schrodinger equation with a linear and cubic potential.
We can further suppose $A(\xi,\tau)= a(\xi,\tau)e^{i \theta(\xi,\tau)}$ which gives us two coupled equations
$$a_{\tau}-2\beta a_{\xi}\theta_{\xi}-\beta a \theta_{\xi\xi}=0$$
$$\rho a -\alpha a^3+\beta a \theta_{\xi}^2- a \theta_{\tau}-\beta a_{\xi\xi}=0$$
which we can slightly rearrange (we divide the second one with $a$ and differentiate wrt $\xi$)
$$(a^2)_{\tau}-2\beta(a^2\theta_{\xi})_{\xi}=0$$
$$(\theta_{\xi})_{\tau}-2\beta(\theta_{\xi})(\theta_{\xi})_{\xi}=-\alpha(a^2)_{\xi}-\beta (a_{\xi\xi}/a)_{\xi}$$
This strongly suggests putting $\Theta=-2\beta\theta_{\xi}$:
$$(a^2)_{\tau}+(a^2\Theta)_{\xi}=0$$
$$\Theta_{\tau}+\Theta\Theta_{\xi}=-2\alpha\beta(a^2)_{\xi}+2\beta^2 (a_{\xi\xi}/a)_{\xi}$$.
The local frequency and wavenumber are described by $\omega=\omega_0+\phi_t$, and $k=k_0-\phi_{\xi}=k_0+\frac{1}{2\beta}\Theta$. If we consider small modulations we can safely discard the $(a_{\xi\xi}/a)_{\xi}$ term. Still, we see that frequency and amplitude modulations are not independent from each other.
We can linearize the equations near harmonic solutions with $a=a_0$ and $\Theta=0$ so that $a=a_0+\tilde a$ and $\Theta=\tilde\Theta$ where $|\tilde a|<<a_0$, $|\tilde\Theta|<<1$. Plugging it in:
$$a_0\tilde\Theta_{\xi}-2\tilde a_{\tau}=0$$
$$\tilde\Theta_{\tau}+4\alpha\beta a_0 \tilde a_{\xi}=0$$
We try solutions of the system with $\tilde\Theta,\tilde a$ proportional to $\exp[i(\Omega\tau-\kappa\xi)]$ and solve the resulting linear system of equations for the amplitudes $\tilde \Theta_0,\tilde a_0$:
$$-\kappa a_0 \tilde \Theta_0 - 2 \Omega \tilde a_0 = 0 $$
$$\Omega \tilde \Theta_0 - 4 \alpha\beta a_0 \kappa \tilde a_0 = 0 $$
and the vanishing of the determinant gives a dispersion relation:
$$\Omega^2=2\alpha\beta a_0^2\kappa^2$$
Finally, we see that we are dealing with either stable or unstable solutions, depending on the sign of $\alpha\beta$, which is the product of the dispersive term $\beta=c_g^2-c^2=\frac{1}{2}\frac{d^2\omega}{dk^2}\big|_{k=k_0}$ and the nonlinear cubic coupling term $\alpha=\lambda/(2\omega_0)$. If the product is positive, we are dealing with stable waves. This is essentially the case with the Jacobi elliptic waves we encoutered earlier, where we wanted $\tilde\lambda=\lambda/(c^2-1)$ to be negative. The notation is a bit off, as there $c$ was our $c_g$, and $c=1$.
On the other hand, for $\alpha\beta<0$ we have imaginary frequencies, and solutions $\exp{\pm \Omega \tau}$, which explodes as $t=\tau\rightarrow\infty$! Here the modulated wave interacts with the carrier wave and instability occurs.
Unfortunately I doubt I could say more about this wave equation! Perturbative solutions in $\lambda$ aren't that viable due to the cubic nonlinearity. The zeroth term would just reproduce an ordinary travelling wave, while the first term would again be complicated. I hope this was helpful!
