The output of this differential amplifier, if it had no constraints on the range of potentials it could actually produce, is:
$$ V_{OUT} = A(V_X - V_Y) $$
Here \$A\$ is gain, which in your case is \$A=\frac{100{\rm k\Omega}}{100{\rm k\Omega}} = 1\$. \$V_X\$ and \$V_Y\$ are the potentials at the ends of the transformer's secondary winding. Forget for the moment what those actual potentials are, just understand that the output is a function of the difference between them only.
The problem you have is that there is a constraint on the output range - set by the op-amp's supply potentials. If that difference is, say, \$V_X - V_Y = -1{\rm V}\$, how on earth is that op-amp supposed to produce an output \$V_{OUT}=-1{\rm V}\$ if its two supply potentials are ground (0V) and +5V? It cannot, so you must either provide the op-amp with a negative supply so that it can actually produce a negative output, or bias the system somehow so that the output is also centered somewhere between 0V and +5V.
I presume your question is about the latter. How do we center the output at, say, +2.5V (half way), and what will be the consequences for absolute potentials \$V_X\$ and \$V_Y\$? The solution is to provide a very low impedance source of +2.5V, which you employ instead of ground for the resistor divider at the non-inverting input:
simulate this circuit – Schematic created using CircuitLab
\$V_{REF}\$ becomes an offset for the output, so that the input-output relationship is now:
$$ V_{OUT} = (V_X - V_Y) + V_{REF} $$
This is how you bias the output, but it biases the inputs too, if they are floating, as is the transformer winding. If you do the math, you'll find that the completely isolated transformer winding between X and Y will have its ends shifted to potentials also within the supply range, unless the op-amp output is saturated.
That is how you bias a differential amplifier like this. Your description of adding "a high value resistor somewhere, something, something" is so vague that I cannot possibly imagine what it means. And since the winding is isolated, whether you use the dotted or non-dotted side is irrelevant. All that changes is whether you are measuring \$V_X-V_Y\$ or \$V_Y-V_X\$.
Just for completeness, it's very unlikely that you have a low impedance source of +2.5V just lying around, so in practice you would probably employ some equivalent, using Thevenin's theorem and the existing supplies of 0V and +5V:
simulate this circuit
Those two are functionally identical, and the version on the right looks suspiciously like R6 and \$V_{REF}\$ in my schematic above. Therefore you can replace R6 and \$V_{REF}\$ with the the left-hand version:
simulate this circuit
$$
\begin{aligned}
V_{OUT} &= (V_X - V_Y) + 2.5{\rm V} \\ \\
\end{aligned}
$$
The question paraphrased is "what are absolute potentials \$V_X\$ and \$V_Y\$?". Here's how you might go about answering that. Start with the basic circuit, with no supply constraints to limit output voltage range:
simulate this circuit
Thousands of times before us engineers have already calculated the relationship between \$V_{OUT}\$, \$V_X\$ and \$V_Y\$. Assuming \$\frac{R_6}{R_5}=\frac{R_8}{R_7}\$:
$$
\begin{aligned}
V_{OUT} &= \frac{R_6}{R_5}(V_X-V_Y) + V_{REF}\\ \\
&= V_X-V_Y + V_{REF} \\ \\
\end{aligned}
$$
We'll treat the op-amp as a voltage source producing \$V_{OUT}\$ according to that relationship:
simulate this circuit
That's a bit hard to follow still, so I'll redraw it here, rearranged for ease of reading:
simulate this circuit
There are some resistors in series that can be combined into single-resistor equivalents:
simulate this circuit
That's pretty easy to solve for \$V_X\$ and \$V_Y\$, if you apply KCL, KVL and Ohm's law. This might not be the easiest way, but it is rigourous:
$$
\begin{aligned}
V_{OUT} - V_J + V_{IN} - V_K - V_{REF} &= 0 \\ \\
V_{OUT} - IR_J + V_{IN} - IR_K - V_{REF} &= 0 \\ \\
V_{OUT} + V_{IN} - V_{REF} &= IR_J + IR_K \\ \\
I(R_J + R_K) &= V_{OUT} + V_{IN} - V_{REF} \\ \\
I &= \frac{V_{OUT} + V_{IN} - V_{REF}}{R_J + R_K} \\ \\
V_{OUT} &= V_X - V_Y + V_{REF}\\ \\
&= V_{IN} + V_{REF} \\ \\
I &= \frac{V_{IN} + V_{REF} + V_{IN} - V_{REF}}{R_J + R_K} \\ \\
&= \frac{2V_{IN}}{R_J + R_K} \\ \\
\end{aligned}
$$
With a value for current \$I\$, we can now find absolute potentials \$V_X\$ and \$V_Y\$ (refer to the schematic to see how KVL yields these next two relationships):
$$
\begin{aligned}
V_X &= V_{REF} + IR_K \\ \\
V_Y &= V_X - V_{IN} \\ \\
\end{aligned}
$$
Of course this only works if the op-amp is not saturated. Once the output gets stuck near a supply potential, and can't extend any further, the relationship \$V_{OUT}=V_{IN}+V_{REF}\$ will no longer be true. The algebra here will therefore only be valid for \$V_{IN}\$ where the corresponding \$V_{OUT}\$ lies between 0V and +5V, or however close the the op-amp output can get to those extremes.
All this maths is done for you by simulators, and while it's not hard to do by hand, as I did here, it's probably much, much simpler to simulate the design and verify that you get the \$V_{OUT}\$ you want for all expected \$V_{IN}\$.
There's one more caveat that we didn't address, and that's what happens at P and Q. These are the op-amp's direct inputs, which have constraints on their acceptable range too, detailed in the op-amp's datasheet. It's equally important to ensure that \$V_P\$ and \$V_Q\$ never fall outside those limits, for which more algebra is required, and which simulators can do much more easily than I could.
So simulate, that's the short answer, and be thankful you don't have to do this algebra every time, like people had to not so long ago.
