I think that you're basically right when you make the following suggestion:
Does part of the input force (or pressure) get wasted in increasing the temperature of the gas and therefore making it less effective?
Recall from the First Law of thermodynamics,
\begin{align}
\Delta E = Q-W
\end{align}
where $\Delta E$ is the change in internal energy of the fluid, $W$ is the work done by the fluid, and $Q$ is the heat transferred to the fluid during the lifting process. Let's assume that the fluid is thermally insulated so that $Q=0$, then we have
\begin{align}
\Delta E = -W
\end{align}
The $W$ term will then be a combination of $W_1$ and $W_2$ where $W_1$ is the magnitude of the work you do to compress the fluid, and $W_2$ is the magnitude of the work done by the fluid in raising the object. In fact, we have
\begin{align}
W = W_2-W_1
\end{align}
so that
\begin{align}
\Delta E = W_1 - W_2.
\end{align}
Now, for an incompressible liquid, no work needs to be done to compress the fluid, so we have $W_1=W_2$, and therefore $\Delta E = 0$; the internal energy of the fluid remains the same. However, as suggested by John Rennie, in the case of a gas, some work goes into compressing the gas, and we have $W_2 < W_1$, so we have
\begin{align}
\Delta E >0
\end{align}
But the internal energy of a (ideal) gas is proportional to its temperature, so this means that its temperature increases a bit during the lifting process.
