Visualize the op-amp as an inverting stage. The - input is a virtual earth (almost: it is off 0V because of the small voltage at the + terminal). Displacement notwithstanding, it is a mixing point. From the point of view of the - terminal, the current source is just an extra input which feeds the mixing point. Because the mixing point is a voltage source, R1 and the current source isolated from affecting each other.
Let's draw a schematic for this concept.
The current that drawn out of the mixing point at the - terminal is pumped into R3, but that is irrelevant and we can separate that by splitting the current source into two.
If the current sources are removed, the behavior of the circuit is very clear. The + terminal is grounded, no current flows across R1 because both ends are at 0V, and the output is 0V.
Next, we can think about happens if the two current sources are added one at a time.
When the second current source, the one feeding R3, is introduced, the effect is that the mixing point is lifted from 0V. Because that happens, there is now a potential difference across R1, since its other end is grounded, and a current flows. The same current flows across R2, creating a potential difference there, which gives us \$V_o\$.
Then, when the current source in parallel with R1 is introduced, there is an additional current flow at the mixing point. This additional current does not disturb the voltage of the mixing point. It simply adds to the current that flows across R2, which adds to the R2 potential drop. We can simply adjust the previous \$V_o\$ with that drop.
R1 is relevant with regard to the displacement of the + terminal voltage.
R1's top is pinned to ground, and so there is a current flow from its bottom which contributes to the flow across R2. If we remove R1, we remove that current and the output changes. R1 is not relevant to the effect of the top current source, because it is in parallel with it. That source adds a certain amount of current to R2, and therefore adds to the voltage drop, regardless of how much R2 current already comes from R1.