Operation of the Ford–Fulkerson algorithm given an almost maximum flow I have been assigned the question:

Let $G$ be a flow network and $f^*$ be the maximum flow computed by the Ford-Fulkerson algorithm. Consider a new flow network $G'$ constructed by increasing the capacity of exactly one of $G$'s edges by one.
Using the maximum-flow and minimum-cut theorem, argue that if we run the Ford-Fulkerson algorithm on $G'$ with $f^*$ as the initial flow, the while-loop shall terminate in at most one iteration.

But I feel like the loop can run more than once...
Considering a base case where $G$ has one edge between the sink and source with a capacity of 1.
Then $G'$ will have one edge with a capacity of 2.
So then when we run the Ford–Fulkerson algorithm with an initial flow of $f^* = 1$, the first loop will leave a residual graph with an edge with a capacity of 1, and therefore the loop will run for a second time.
You are interpreting the question too literally. The question claims that only one augmentation step will be performed. In your example, we augment $f^*$ by pushing one more unit of flow, and then we reach the maximum flow of the updated network.