If $\Gamma$ is a discrete subgroup then the answer to your question is yes. Starting with a hyperbolic structure from a regular octagon with $\pi/4$ angles as you describe, consider the map from the octagon to the quotient surface $\Sigma = \mathbb{H}^2 / \Gamma$. The boundary of the octagon maps to a 1-complex consisting of four based closed geodesics $\hat\alpha,\hat\beta,\hat\gamma,\hat\delta$, all with the same base point $p \in \Sigma$ but otherwise disjoint. Each of these geodesics has a $\pi/2$ angle at $p$, though. If you pull them tight, you get four totally geodesic embedded circles $\alpha, \beta, \gamma, \delta$. The circles $\alpha,\beta$ intersect each other transversely in 1 point, as do the circles $\gamma,\delta$, and the $\alpha,\beta$ pair is disjoint from the $\gamma,\delta$ pair. The complement $S - (\alpha\cup\beta\cup\gamma\cup\delta)$ is an annulus with ``scalloped'' boundary, each boundary component a concatenation of 4 geodesic arcs; the surface $S$ can be reconstructed by gluing these arcs in pairs in the appropriate fashion. Now pick any hyperbolic structure on $\Sigma$, that is, any discrete faithful representation $\Gamma \to \textrm{Isom}^{+}(\mathbb{H}^{2})$. All nontrivial elements are hyperbolic. When you straighten the four circles $\alpha,\beta,\gamma,\delta$ in this new hyperbolic structure, you get the same intersection pattern and the same qualitative description of the complement, however the "shape" of that annulus has changed, i.e. the length of the core curve, the lengths of and angles between the scalloped sides, and possible "twisting" around the core curve. But you can still "pull in" the boundary of this annulus to give the octagon structure that you ask for. To do this, pick a corner on each boundary circle, drag that corner to the core circle of the annulus, and then drag the two corners along the core circle until they touch; when you do this dragging maneuver, no new identifications of the annulus boundary will be introduced until the two points touch and the annulus becomes an octagon with side pairings as you ask for (maybe you'll have to drag around and around the core curve in order to get the exact side pairings, rather than getting them conjugated by a power of the core curve due to twisting around the core of the annulus).

On the other hand, I believe there do exist injective homomorphisms $\Gamma \to \textrm{Isom}^{+}(\mathbb{H}^{2})$ with nondiscrete image so that $a,b,c,d$ are all hyperbolic, in which case what you ask for is hopeless. This kind of stuff can be found, I believe, in Bill Goldman's thesis, but I do not know a published reference. For an example, pick one hyperbolic element of $\textrm{Isom}^{+}(\mathbb{H}^{2})$ and map all of $a,b,c,d$ to that element, which induces a non-injective homomorphism $i : \Gamma \to \textrm{Isom}^{+}(\mathbb{H}^{2})$. It is possible to perturb $i$ to become injective. But the Euler number of a representation is continuous under this perturbation, and the representation $i$ itself has Euler number zero, so the perturbed representation does too. It must therefore be indiscrete, because discrete faithful representations of $\Gamma$ all have Euler number $\pm 2$.