How am I able to stand up and walk down the aisle of a flying passenger jet?

Due to momentum being conserved, when you accelerate yourself forwards relative to the plane, the tangential force you're applying to the floor will accelerate the rest of the plane backwards. Since the plane has a lot more mass than you, its velocity will not change by very much.

Thus, an inertial observer who was initially at rest with respect to the plane (and you) will see both you and the plane gain kinetic energy (due to your muscle work). The vast majority of the additional kinetic energy goes into you, though.

However, an observer on the ground will see the rest-of-the-plane slow down slightly, which means that it loses quite a bit of kinetic energy due to its large mass and velocity. This loss of kinetic energy from the plane cancels out the additional kinetic energy the ground observer thinks you gained, so the ground observer's energy ledger still balances.

(Mathematically, to the ground observer $v$ is, to a first approximation, both the ratio between your your gained momentum and your gained kinetic energy, and the ratio between the plane's lost momentum and its lost kinetic energy. So conservation of momentum leads to conservation of total energy, to first order. The term that comes from your muscle work is a second-order effect).

Both observers agree on the amount of energy your muscles contribute (at least as long as relativistic effects can be ignored).

Kinetic energy is not invariant under Galilean transformations. To see this consider the following:

In the rest frame of the plane you apply a force $F$ of 100N for one second to accelerate yourself to 1 m/s. During this time you move a distance $d$ of 0.5m so the work done is:

$$ W = Fd = 100 \times 0.5 = 50\,\text{J} $$

This of course is equal to your kinetic energy of:

$$ E = \tfrac{1}{2}mv^2 = \tfrac{1}{2} \times 100 \times 1^2 = 50\,\text{J} $$

The observer on the ground sees you applying a force of 100N for one second, but because the plane is moving at 250m/s the ground observer sees you move a distance of 250.5m. Therefore the work done is:

$$ W = Fd = 100 \times 250.5 = 25050\,\text{J} $$

For the ground observer your initial KE, before you started walking, is:

$$ E = \tfrac{1}{2}mv^2 = \tfrac{1}{2} \times 100 \times 250^2 = 3125000\,\text{J} $$

And your kinetic energy after you've reached a speed of 1m/s is:

$$ E = \tfrac{1}{2}mv^2 = \tfrac{1}{2} \times 100 \times 251^2 = 3150050\,\text{J} $$

So the change in your kinetic energy is:

$$ \Delta KE = 3150050 - 3125000 = 25050\,\text{J} $$

And as before this is equal to the work done.

Response to comment:

user2800708 points out, quite reasonably, that your muscles only produced 50J so if the ground observer sees your kinetic energy change by 25050J where did the rest of the energy come from?

The answer is that when you propel yourself forwards with a force of 100N you propel the plane backwards with a force of 100N. So in order to keep its speed constant at 250 m/s the planes engines have to provide an extra 100N of thrust. In the one second we watch the plane it moves 250m, so the extra work done by the plane's engines is:

$$ W_\text{plane} = Fd = 100 \times 250 = 25000\,\text{J} $$

Add this to the 50J provided by your muscles and we get the 25050J that we calculated above.