Problems that depict situations where the tensions are same on ropes on both sides of the pulley are ideal situations.It is stated so in order to minimize any complexities that may arise if the pulley was to rotate.Now, if the tensions were not equal on both sides, the pulley would experience a net non-zero torque and hence a net angular acceleration and eventually rotate.Also,these are cases where pulleys have friction between its rim and the rope..if there were no friction on the rope-rim interface the pulley would not turn and its mass would become irrelevant . 
When you have a situation like you describe, with block A on the left and block B on the right, then the tension in A will cause a counter-clockwise torque, and the tension from B will cause a clockwise torque. Now you need to take this direction into account exactly once. So either:
- you subtract the two tensions, and then apply a net tension with a clockwise torque, OR
- you say that one tension applies a clockwise torque, and the other an anti-clockwise torque
The net torque in the first case is
$$\Gamma_{net}=(T_a-T_b)\cdot R$$
and in the second case:
$$\Gamma_{net} = T_a \cdot R - T_b \cdot R$$
as you can see, these are the same.
From your description, it sounds like you are doing the right thing; let me just write out how I would tackle it from here - let's see if I get a different answer than you.
Using the conventions that is A on the left, B on the right, acceleration is positive when A moves down (it is heavier), and torque is counterclockwise. Then
$$\begin{align}T_a &= m_a(g-a)\\
T_b &= m_b(g+a)\\
\Gamma &= (T_a-T_b)\cdot R \\
&= (m_a(g-a) - m_b(g+a))\cdot R\end{align}$$
$$\Gamma = ((m_a - m_b) g - (m_a + m_b) a)\cdot R\tag1$$
Now we know that this torque is what accelerates the pulley which has moment of inertia $I$:
$$\Gamma = I \dot \omega = I \frac{a}{R}\tag2$$
Combining these two equations,
$$I \frac{a}{R}=((m_a - m_b) g - (m_a + m_b) a)\cdot R\\
a\left(\frac{I}{R} + (m_a + m_b)R\right) = (m_a - m_b) g\cdot R\\
a = \frac{(m_a-m_b)gR^2}{I+(m_a+m_b)R^2}=\frac{(4.5-2.0)*9.8*0.14^2}{0.4 + (4.5+2.0)*0.14^2}=0.91\; \rm{m/s^2}$$
Is that not what you are getting?
Best Answer
Using the principle of virtual work, if you move the blocks a distance a, the inclined block is lowered by an amount equal to $a\sin(\theta)$, meaning that it gains energy $m_2 ga\sin(\theta)$. The total moving mass is $m_1 + m_2$, so that the acceleration is the same as for a mass $m_1 + m_2$ in 1 dimension with a force $m_2 g \sin(\theta)$, so that
$$ a = {m_2 \over m_1 + m_2} g \sin\theta $$
This is how you solve these types of problems, it's equivalent to writing the Lagrangian, but more elementary sounding.