How does work from other sources affect Gibbs free energy of the system?
I will treat the first and second question as if they were part one and part two of a single question. I think the OP wants to know how non-PV work plays a role when looking at the Gibbs energy of a reaction, and its interpretation.
To start off, the Gibbs energy for a given reaction is a state function, so it does not depend on how much work or how much heat are exchanged between system and surrounding. The total change in entropy, however, is affected by what is going on in the system and in the surrounding, so work vs heat is relevant for the second law of thermodynamics and the direction of the reaction.

First Question: In the case of an electrolytic cell, in the system, there is also a power source. How would this external voltage source affect the Gibbs free energy equation? Would it just be an extra $-VIt$ term or something else which would cause $\Delta_\mathrm rG &lt; 0?$

The easiest way to deal with the power source is to place the system/surrounding boundary such that the power source is in the surrounding. This shows it does not affect the Gibbs energy of the reaction at all. The Gibbs energy is positive, but the reaction can go forward because it is offset by the work done on the system. As a result, the combined entropy increases, and we are not breaking the second law.
In general, the criterion is not that the Gibbs energy of reaction has to be negative, but that
$$\Delta_r G &lt; w$$
When there is no work done on the system or by the system, this simplifies to the requirement that the Gibbs energy be negative.
As an aside, when the Gibbs energy of reaction is negative, we can not have the reaction do an arbitrary amount of work on the surrounding, but that amount is limited to the Gibbs energy. That's why the Gibbs energy is sometimes called the maximal work (this is a good term for cases when it is negative; when it is positive, it is the minimal work that needs to be done on the system for it to be possible for the reaction to go forward).

Second Question: In the case of a biological cell, the work usually comes from the hydrolysis of ATP. How would this alter the equation and cause $\Delta_\mathrm rG &lt; 0?$

The hydrolysis of ATP usually does not do work on the surroundings. Instead, the uncatalyzed reaction has a positive Gibbs energy of reaction. To use up the reactants and make the products, the cell runs a different reaction that includes the hydrolysis of ATP. It is not sufficient that ATP be hydrolized somewhere while the desired reaction goes on somewhere else. Instead, the two reactions are coupled so that ATP hydrolysis can not occur without the other part of the reaction happening. Here is an example how that might look for the reaction $\ce{A + B -&gt; C}$:
$$\ce{A + ATP -&gt; A-P + ADP}\tag{1}$$
$$\ce{A-P + B + H2O -&gt; C + P_i}\tag{2}$$
In this scheme, A-P refers to a phosphate group bound to A, and $\ce{P_i}$ refers to phosphate. Both steps can have a negative Gibbs energy of reaction, even thought the Gibbs free energy of reaction for $\ce{A + B -&gt; C}$ is positive. As net result, A and B turn in to C, without work done on the system.
There are other cases where there is work done on the system or by the system. These include proton pumps, and myosin hydrolyzing ATP to contract muscle fibers. You could also discuss photosynthesis in terms of work done to drive the synthesis of ATP.