Will this this schematic modification work?

The original circuit has the indicator LED wired in such a way that could destroy it, and while the overall design could work, it's certainly got problems. Your replacement of that LED with R1 (I think you should have called it R3, since the R1 designator is already used for a different resistor) does solve the LED problem, and might actually work, if these two conditions are met:

1. The value of R1 is correctly chosen to develop an appropriate voltage across it given whatever range of photo-current you expect from the photo-diode. I'll address this in a moment.

2. The GPIO input is an ADC, measuring an analogue voltage. The "output" of this system is not a digital "high-or-low" potential, rather it will be a continuously varying analogue potential lying somewhere between 0V and +3.8V or so (assuming your supply is 5V), requiring an analogue-to-digital converter to "measure" it.

The new 510Ω resistance is way smaller than it needs to be, and while its exact value is not critical, the system will draw much less current if you use something like 10kΩ instead. This will also reduce the amount of photo-current diverted to the darlington-pair emitter-follower, leaving as much current as possible for R1 (the original R1, not your new one) to convert to a measurable voltage.

The best way to know what value you should choose for that R1 is to study the datasheet for your photo-diode, which will give you figures for photo-current as a function of illumination. That's not easy, because you have to consider "dark current", which is present due to the always-non-zero voltage across the photo-diode. The good news is that you can find a value for R1 by trial and error, which maybe your best approach if you're not very familiar with photo-diodes, and their behaviour. Here's a procedure you can use to find an appropriate value for R1. Start with a simple test setup, consisting of the 5V supply, the photo-diode D1, and resistor R1, and a digital voltmeter:

^{simulate this circuit – Schematic created using CircuitLab}

$$ V_W = I_D \times R_1 $$

Illuminate your photo-diode with as much light as it will ever experience, the most illuminated it can be in your intended environment, with the transmitter LED fully lit. Then install different values for R1 until the voltmeter reads something in the vicinity of 4V. This will give you some room for maneuver as light level varies, but this will be at the upper end of the illumination scale.

This value is potential \$V_W\$ that will be applied to the base of the darlington-pair formed by Q1 and Q2. They are configured as an emitter follower, whose emitter potential will be about 1.2V lower:

^{simulate this circuit}

$$ V_X = V_W - 1.2{\rm V} $$

\$V_X\$ is the potential you will measure using one of the Arduino's ADCs. It cannot rise (significantly) above +3.8V, or fall below 0V, since these limits are imposed by the 5V supply you use to power D1/R1, and the 1.2V "drop" due to the transistors. That +3.8V upper bound is too close to the Arduino's high/low switching threshold, which is why you can't reliably use this as a digital signal operating a digital GPIO Arduino input. Without further modification, you should to use an analogue ADC input to measure this potential, not a digital input.

^{simulate this circuit}

That's how you draw a schematic - signals flowing left to right (where possible), higher potentials (+5V, +9V) at the top, lower potentials (0V) at the bottom. A clearly labelled ground node to represent what you are going to call "zero-volts", and clear separation of function into obvious modular parts; the photo diode and its resistor clearly distinct from the amplifier stage, which is clearly separated from the Arduino and power systems.

This should work, but as I explained, everything depends on R1, and the environment - whatever value of R1 you choose for a brightly sunlit room will be completely useless on a cloudy day.

Both circuits are not good. The two transistors are connected as common collector, that means they provide zero voltage amplification. So the Mcu will see a very small signal change between dark and light.

You need to use a common emitter amplifier that gives you high voltage gain and has high input impedance.

After leaving a few comments, I think I have enough for an answer.

The raw circuit that you copied from has some problems, related to not having any kind of current limit for the indicator LED. The transistors enforce the same voltage across the LED as there is across R1 (minus 0.7V for each or 1.4V total), using whatever current is necessary to do that. So it's effectively using the battery itself as the indicator's current-limiting resistor.

That will probably drain the battery rather quickly, and it'll affect the rest of the circuit too, by dropping the supply voltage. And depending on the internal resistance of the battery, it may even kill the indicator LED and/or Q1.

Your modification actually fixes all that, by replacing that LED with a resistor. But with the way you're actually using it, it's way overkill.

Just delete the transistors, and run the junction between the photodiode and its resistor (the original R1) directly into an analog input. You may need to tweak the value of R1 to have the signal sit somewhere "nice" (whatever that means for your application), and then you can apply whatever threshold / hysteresis / timing / frequency filtering / other logic you want in software.