Heat pump should transfer heat from outside into the house. It should not generate heat (ideally) it should only force the heat to move.
One joule of work executed by the heat pump can transfer several joules of heat - for example 4 joules (it is the reason why the heat pump is efficient "source" of thermal energy). This ratio is called Coefficient of Performance or COP.
The bigger is the difference between input and output temperature -> the more work must the heat pump do to transfer the heat. Therefore the COP is decreasing.
If the COP did not decrease with increasing temperature difference, it would be possible to construct a perpetual motion machine of the second kind.
If work required to compress the gas was included into the computation the COP would be visible. And in order to achieve good COP the much lower output temperature would be needed.
EDIT: Computation of $COP$ with example.
We will start from ideal heat engine modeled by Carnot cycle. Carnot cycle has efficiency $$\eta = \frac{T_H - T_C}{T_H}$$
Let's assume $T_H=310K$, $T_C=270K$ and assume $100J$ of heat $Q$ will be delivered from hot reservoir to the heat engine.
As a result $$\eta \times 100J = \frac{310 - 270}{310} \times 100J = 12.9J$$ of work $W$ will be done by heat engine and $87.1J$ of heat will end up in cold reservoir.
What happens when we reverse the process? (Carnot cycle is reversible)
In the reversed process $87.1J$ of heat will be taken from cold reservoir, $12.9J$ of work will be delivered to the engine (now the heat pump) and $100J$ of heat will end up in the hot reservoir.
The COP is $$COP = \frac{Q}{W} = \frac{100J}{12.9J} = 7.75$$
And in general $$COP_{ideal} = \frac{Q}{W} = \frac{Q}{\eta \times Q} = \frac{1}{\eta} = \frac{T_H}{T_H - T_C}$$
Look up something called the Carnot efficiency. That is the theoretical limit of how effecient any heat engine can be at converting heat power to some other form. This maximum possible efficiency is
Carnot efficiency = Tdiff / Thot = (Thot - Tcold) / Thot
By simple 8th grade algebra, you can see that you get a higher value by decreasing Tcold (the cold side temperature) than by increasing Thot (the hot side temperature) by the same amount.
For example, the Carnot efficiency of 100°C to 0°C is 100°K / 373°K = 26.8%. Adding 10 degrees to the hot side you get 110°K / 383°K = 28.7%, but decreasing the cold side by the same 10 degrees yields 110°K / 373°K = 29.5%.
Best Answer
It's pretty much as you say. But it sounds as though you're trying to guess. I'd suggest a couple of little sketches and some first principles reasoning; see my drawing below
Both the circles are my reversible heat engine. On the left, it is running "forwards" and yielding work $W$ from the nett flow $Q_C$ into the cold reservoir at temperature $T_C$ after extracting heat $Q_H$ from the hot reservoir at temperature $T_C$. On the right it has work $W$ done on it for a nett flow nett flow $Q_C$ out from the cold reservoir, and the total $Q_H=Q+Q_C$ is pumped into the hot reservoir. By definition of the thermodynamic temperature in terms of heat flow ratios in reversible heat engines, as I discuss in my answer here, then
$$\frac{Q_H}{T_H}=\frac{Q_C}{T_C}\tag{1}$$
and, by energy balance
$$Q_H = W+Q_C\tag{2}$$
The engine is reversible, so that all the magnitudes of the energy quantities are the same; their signs are opposite: instead of the heat engine doing work $W=Q_H-Q_C$, we must do work on the heat engine to pump $Q_C$ from the cold reservoir and thereafter dump the total energy $Q_H=W+Q_C$ back into the hot reservoir.
So you now work out both efficiencies from (1) and (2): for the heat pump, a sensible "efficiency" is the fraction of the heat drawn from the hot reservoir that we convert into work; from (1) and (2) it is:
$$\eta = 1-\frac{T_C}{T_H}$$
For a heat pump, a sensible definition of "efficiency" is how much heat $Q_C$ we draw out of the cold reservoir for input of unit work: this ratio is:
$$\frac{Q_C}{W} = \frac{Q_C}{Q_H-Q_C} = \frac{T_C}{T_H-T_C}$$
although this one is a bit weird insofar that it can be greater than 1. Note that it becomes infinite as $T_C\to T_H$: heat can move spontaneously without work input between two reservoirs of the same temperature (although for a finite reservoir, the flow would lead to a temperature difference and halt the flow). A better word, which I believe heat pump engineers use is "co-efficient of performance": the bigger it is, the less energy you need to put in to remove unit heat from the cold reservoir.