Is it possible to get an example of two matrices

No. If $\operatorname{rank} A < 2$ then either $A = 0$ or $\operatorname{rank} A = 1$, and similarly for $B$. Obviously if $A = 0$ then $$\det(A - \lambda B) = - \lambda \det B = 0$$ since $B$ is not full-rank, and similarly if $B = 0$.

So suppose $\operatorname{rank} A = \operatorname{rank} B = 1$. Then the image of $A$ is some one-dimensional space $\operatorname{span} \{ v\}$ and the image of $B$ is some one-dimensional space $\operatorname{span} \{ w\}$. It follows that for any $u \in \mathbb{R}^4$ $$(A - \lambda B) u = A u - \lambda B u = c v + d w \in \operatorname{span}\{v,w\}$$ for some $c, d$. In particular, $\operatorname{rank} (A - \lambda B)$ is at most two, and thus $\det (A - \lambda B)$ is always zero.

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If you meant to say $\operatorname{rank} A \leq 2$ and $\operatorname{rank} B \leq 2$ instead, then the answer is yes; just take $$A = \begin{pmatrix} 1 \\ & 1 \\ & & 0 \\ & & & 0 \end{pmatrix}, \qquad B = \begin{pmatrix} 0 \\ & 0 \\ & & 1 \\ & & & 1 \end{pmatrix}$$ for instance. In fact, any random pair of rank-two matrices will have this property with 100% probability.

Note that $rank\ A, B < 2$ means their ranks are 1 (If any of them is 0 your question has a trivial answer). That means that all columns of $A$ are $v, \alpha_v v, \beta_v v, \delta_v v$ and the columns of $B$ are $u, \alpha_u u, \beta_u u, \delta_u u$. Now note that the columns of $(A - \lambda B)$ will be sums of those multiples of $u$ with multiples of $v$. Regardless of how you distribute the $v, \alpha_v v, \beta_v v, \delta_v v$ with the $u, \alpha_u u, \beta_u u, \delta_u u$, all columns of $(A - \lambda B)$ will be of the form $xu + yv$ for some $x$ and $y$ scalars. Therefore, all your 4 columns will be a linear combination of the two vectors $u$ and $v$ and therefore the rank of $(A - \lambda B)$ can never be greater than 2.