Theoretical Refridgeration
The coefficient of performance metric for the fridge is
$$CoP_R = \dot{q}_c/\dot{w}_R = T_c/(T_h - T_c)$$
Heat $\dot{q}_c$ is removed from the fridge with power $\dot{w}_R$. For an ideal case, this is defined by the temperature in the fridge $T_c$ and the temperature at the external coils $T_h$. You want to keep $T_c$ as low as possible. You control $\dot{w}_R$ through the electrical power to the fridge $P$ and the (irreversible) efficiency of the motor $\epsilon_M$ as $\dot{w}_R = \epsilon P$. The system defines $\dot{q}_c$ as the heat flow that enters the fridge from the outside. You must remove at least that to sustain the temperature $T_c$. The temperature $T_h$ is a design setting for the compressor and fluid that are used in the cooling coils themselves. Finally, $\dot{q}_c$ is related to the heat that must be pulled from the coils $\dot{q}_h$ by the expression $\dot{q}_c + \dot{w}_R = \dot{q}_h$. Heat transfer into the fridge and away from the coils are addressed as theoretical problems in heat transfer.
Theoretical Heat Transfer
The heat flow into the fridge $\dot{q}_i$ has to be removed as $\dot{q}_c$ by the heat pump (cooling cycle). How can it be decreased? Consider it from heat transfer as below.
$$ \dot{q}_i = U A (T_a - T_c) $$
Here, $U$ is the overall heat transfer coefficient for the system, $A$ is the essentially external area of the fridge, and $T_a$ is the surrounding air temperature. We can expand $U$ as a sum of resistors for external convection $R_a = 1/h_a A$, wall conduction $R_w = w/kA$, and internal convection $R_c = 1/h_cA$. In these expressions, $h_j$ is the air or internal convection coefficient, $w$ is the wall thickness, and $k$ is the thermal conduction of the wall.
At the back side, convection around the coil to the surrounding air pulls out the heat that is dumped to the air. This affects the temperature of the coil as below.
$$ \dot{q}_h = h_a A_c (T_h - T_a) $$
In this, $A_c$ is the area of the coils.
Practical Insights
Let us assume that you have no control on $T_a$.
As two first steps to lower $T_c$ with all else the same, you can improve the efficiency of your motor to increase $\epsilon$ or you can increase the power to the motor to increase $P$.
In the heat transfer realm, you can make changes to decrease the heat transfer into the fridge to lower $\dot{q}_i$. Ideally, you would want to DECREASE the convection around and inside the fridge to lower $h_a$ and $h_i$. This means that a fridge sitting in stagnate air is the best case. Alternatively, you can increase the thickness of the wall to the fridge $w$ or decrease the thermal conductivity of the insulation in the walls $k$.
The opposing case is to make changes that will increase $\dot{q}_h$. Here, the immediate inverse recommendation arises to INCREASE $h_a$ as convection around the coils or to increase the area of the coils $A_c$.
Practical Suggestions
One suggestion is to change the insulation in your fridge to a material with a higher $R$ value. This will decrease $U$ and lower the heat flow into the fridge from the outside.
Another suggestions is to work on increasing the convection flow around the coils. Ask the hotels to have holes drilled in the bottom back of the box. This will improve the natural convective flow of air along the coils at the back of the fridge. The wider the holes or the opening is for air flow, the better. The best option is when the entire bottom of the box is a mesh that is mechanically strong enough to hold the weight of the fridge yet open enough to allow full air flow from below the fridge. This will only work when the box also includes a wide enough hole at the top to vent the hot air.
Absent the option to change the box design, design a channel system attachment to the fridge. The channel should allow natural convection to pull cooler air from the front along the bottom, run it up the coils in the back, and vent it out the back of the box.
An alternative answer is to redesign the fridge to include cooling fans to blow the air. The power could be pulled from a part of the power to the fridge itself. This will be a better cooling system than the passive one above. The downside is, this may generate more noise than a hotel will tolerate.
Other Thoughts
When the hotel has their specifications on cooling demand, you should set your specifications on ventilation to meet that cooling demand. Otherwise it seems that you put yourself in a box that you allow them to dictate unilaterally that your system must meet their criteria regardless of where they put it. I know of no heat transfer system that functions the same always independently of its surroundings.
