This is a correction of my original answer, where I misread the matrix $b$, I somehow read that it has three distinct eigenvalues, not two.
First of all, there is no "the representation $\rho$ with $\rho(A), \rho(B)$ conjugate to above matrices": There is a continuum of such representations, even taken modulo conjugation. One can ask, however, the following questions:
Is it true that some representations $\rho$ with $\rho(A), \rho(B)$ conjugate to above matrices $a, b$ are reducible? This question has positive answer by taking a representation with image in the subgroup of diagonal matrices (just take $\rho(A)$, $\rho(B)$ diagonal).
Is it true that all representations $\rho$ with $\rho(A), \rho(B)$ conjugate to above matrices $a, b$ are reducible?
The second question also has a positive answer, but the proof is more complicated.
Theorem. Every representation $\rho: \Gamma\to SL(4, {\mathbb C})$ which sends $A$ to a matrix conjugate to $a$ and sends $B$ to a matrix conjugate to $b$ is reducible: It has an invariant plane in ${\mathbb C}^4$.
Proof. First of all, suppose that $M\in SL(4,{\mathbb C})$ is a diagonalizable matrix with eigenvalues $\lambda, \lambda$, $\mu, \mu$ (listed with their multiplicity), such that $\lambda\ne \mu$. Let $E_\lambda, E_\mu$ denote the eigenspaces of $M$ corresponding to $\lambda$ and $\mu$ respectively. Then the set of $M$-invariant planes in ${\mathbb C}^4$ is a subvariety consisting of three components. Two of them are singletons, $\{E_\lambda\}, \{E_\mu\}$, and the remaining component $C_M$ is a 2-dimensional smooth subvariety isomorphic to $({\mathbb C} P^1)^2$. Elements of $C_M$ are the planes $V\subset {\mathbb C}^4$ intersecting both $E_\lambda, E_\mu$ along lines. The subvariety $C_M$ sits naturally in the Grassmannian $Gr_2({\mathbb C}^4)=G(4,2)$ (2-planes in ${\mathbb C}^4$). Its Poincare dual is a cohomology class $[C_M]^*\in H^4(G(4,2))$.
I will need some basic knowledge of the cohomology ring $H^*(G(4,2))$ conveniently discussed in here.
You can find much more on cohomology groups of Grassmannians elsewhere as well. (Just google "cohomology ring of Grassmannian".)
The Grassmannian $G(4,2)$ is a smooth connected 4-dimensional complex projective variety (all dimensions are complex in what follows). The $H^2(G(4,2))$ is generated by the Schubert class $a_2$ which is Poincare dual to the complex 3-dimensional cycle
$S_P$ consisting of 2-planes which have nonzero intersection with a fixed 2-plane $P\subset {\mathbb C}^4$. (Actually, for the following proof you do not need to know that the dual of $[S_P]$ generates $H^2(G(4,2))$, it suffices to know that the class $[S_P]$ is nonzero (as the fundamental class of any irreducible projective subvariety in a complex projective variety). Then the fact that $H^2(G(4,2))\cong {\mathbb Z}$ will imply that the Poincare dual of $[S_P]$ is a positive multiple of the generator $a_2$.)
The main thing to know is that
$$
a_2^4\ne 0\in H^8(G(4,2)).
$$
(See the link above.) In particular, if $P_1,...,P_4$ are four planes in ${\mathbb C}^4$ then
$$
S_{P_1}\cap S_{P_2}\cap S_{P_3}\cap S_{P_4}\ne \emptyset.
$$
Suppose now that $M_1, M_2$ are two diagonzalizable matrices in $SL(4,C)$ each of which has exactly two distinct eigenvalues $\lambda_1, \mu_1$ and $\lambda_2, \mu_2$, of multiplicity 2 each.
Lemma. The linear transformations represented by $M_1, M_2$ preserve a 2-dimensional subspace $Q$ in ${\mathbb C}^4$.
Proof. Observe that
$$
C_{M_k}= S_{E_{\lambda_k}}\cap S_{E_{\mu_k}}, k=1, 2.
$$
Hence, as we observed above,
$$
C_{M_1}\cap C_{M_2}\ne \emptyset.
$$
Therefore, there exists a plane $Q\subset {\mathbb C}^4$ which belongs to both $C_{M_1}, C_{M_2}$, hence, invariant under both $M_1, M_2$. qed
Specializing to the case $\lambda_1=1, \mu_1=-1$, $\lambda_2=\omega, \mu_2=\omega^2$, where $\omega\ne 1$ is a cube root of $1$, we obtain the assertion of the Theorem. qed
