Restated, does freezing point depression affect the entire amount of
energy needed to melt it (eg. making it melt faster), or does it only
seem to be a noticeable phenomena only at the melting point only?
The entire energy needed to melt the pure cube will be greater than that required to melt the impure cube, and the impure cube will melt faster (start to melt sooner and, when it does melt, at a faster rate). Consider the following:
We start with the following assumptions.
Both cubes have exactly the same mass, surface area, geometry, etc. That is their physical properties are identical and how they physically interact with the surrounding air is identical.
The specific heats $C$ of both the pure and impure cubes are the same.
The latent heats of fusion $h_f$ are the same for both cubes.
The overall heat transfer coefficients, $h_C$ are the same for both cubes.
The air is considered a 33 C thermal reservoir, i.e., its bulk temperature is constant during the heating of the cubes.
Regarding assumptions 2-4 I was not able to locate any data differentiating these properties based on the purity of water. These assumptions may not be correct, but I doubt any differences would be significant enough to alter the conclusions.
We start out with both cubes being at any arbitrary temperature less than the lower of the two melting temperatures, that is, less than 30 C. Up to the point where the temperature of the impure cube reaches 30 C, the rate of heat transfer and the total heat transfer to each cube is the same.
When the two cubes reach 30 C, phase transition begins for the impure cube. The amount of heat required to melt the impure cube, where $h$ is the latent heat of fusion, is
$$Q_{impure}=mh_f$$
The rate at which heat is added is
$$\dot Q_{impure}=h_{c}A(33^{o} C - 30^{o} C)$$
Where $h_c$ is the overall heat transfer coefficient of the ice/water mixture.
During the time melting of the impure cube begins, the air is still raising the temperature of the pure cube. The heat required to bring the pure cube to its melting point of 32 C, is
$$Q_{pure}=mC_{ice}(32-30)^{0}C$$
This heat is over and above that which was required to bring the pure and impure cubes to the melting temperature of the impure cube. This means that, overall, more heat was required to melt the pure cube than the impure cube.
Once the pure cube reaches its melting temperature, the heat required to convert it from ice to water is
$$Q_{pure}=mh_f$$
Since the latent heat is assumed the same as the impure cube, this heat is the same as needed to melt the impure cube. Or $Q_{pure}=Q_{impure}$.
The rate at which heat is added to convert the pure cube ice to water is
$$\dot Q_{pure}=h_{c}A(33^{o} C - 32^{o} C)$$
Where $h_c$ is the overall heat transfer coefficient of the ice/water mixture, assumed to be the same for the pure and impure ice/water phase.
Under the assumption that $h_c$ is the same for the pure and impure cube, the heat transfer rate during the phase transition for the pure cube will be less than the impure cube because of the smaller temperature difference of 1 C for the pure ice vs. 3 C for the impure ice.
Given the above, if the assumptions are correct, I would conclude
More heat is required to melt the pure ice than the impure ice, the difference being the additional heat needed to bring the pure ice to its melting point compared to the overall heat required to melt the impure ice. This additional heat is $$Q_{pure}=mC_{ice}(32-30)^{0}C$$
The impure ice will melt faster/sooner than the pure ice for two reasons: (A) the impure ice will begin to melt before the pure ice since its melting temperature is reached first and (B) during the transition phases (ice to water), the impure ice will melt faster than the pure ice because the temperature difference between the air and the impure ice during melting of 3 C, was greater than the temperature difference between the air and the pure ice during melting of 1 C.
Hope this helps.