So due to Newton's third law there is an equal and opposite force on the ground. Treat this as an object with very large mass $M$, your hockey puck has mass $m$. The puck travels over the ice with speed $v$, the observer moves over the ice in the opposite direction with speed $u$. To this observer the puck seems to have speed $v + u$ and the rink seems to have speed $u$, so the total momentum is $M u + m u + m v.$ The total kinetic energy at the beginning is $$K_0 = \frac 12 M u^2 + \frac12 m (u+v)^2.$$After the slowdown the two have the same velocity $u + \frac{m}{M+m}v$ to conserve momentum, and the total kinetic energy is now $$K_1 = \frac12 (M+m) \left(u + \frac{m}{M+m} v\right)^2.$$
Expanding out we see $$K_1 = \frac12 (M+m) u^2 + m u v + \frac 12 \frac {m^2}{M+m} v^2.$$ On the flip-side, $$K_0 = \frac12 (M + m) u^2 + m u v + \frac 12 m v^2.$$ The difference, $$\Delta K = K_1 - K_0 = -\frac12~ v^2 ~\frac{Mm}{M+m}, $$ is therefore totally independent of $u$.
In general, the kinetic energy of a bunch of particles under a velocity addition is given by $$K' = \sum_i \frac12 m_i (\vec u + \vec v_i)^2,$$
or after expansion,
$$K' = K + \vec u \cdot \vec P + \frac12 M u^2 $$where $M = \sum_i m_i$ and $\vec P = \sum_i m_i v_i.$ Therefore in general, subtracting value 1 from value 0, we find $$\Delta K' = \Delta K + \vec u \cdot \Delta \vec P,$$ and any interaction of these particles which conserves momentum $(\Delta \vec P = 0)$ will cause all observers to agree on changes in kinetic energy, even though they won't agree in general on the specific values of that kinetic energy.
