You've got a number of resonances by others on your question. But I'll dive directly in to demonstrate that it's not an assumption by Tinker.

The Shockley diode model can be simplified for these purposes to the following:

$$I_\text{D}=I_\text{SAT}\cdot\left[\exp\left(\frac{V_\text{D}-R_\text{D}\cdot I_\text{D}}{\eta\,V_T}\right)-1\right]$$

Solving that for \$I_\text{D}\$ would take a page's worth of math steps, which I'll avoid. (Uses branch-0 of LambertW function, which gives \$u\$ for \$y=u\, e^u\$ when you know \$y\$.)

But it can be rewritten in this more easily reckoned with equation:

$$V_\text{D}=I_\text{D}\cdot R_\text{D}+\eta\, V_T\cdot \ln\left(1+\frac{I_\text{D}}{I_\text{SAT}}\right)$$

Apparently, Tinker uses \$\eta=1.75\$, \$R_\text{D}=6\:\Omega\$, and \$I_\text{SAT}=4\times 10^{-21}\$ for its RED LED diode model. And \$V_T=26\:\text{mV}\$ for its thermal voltage.

In your case, with \$1\:\text{k}\Omega\$, you got \$7.04\:\text{mA}\$. So:

(6*x+1.75*.026*ln(1+x/4e-21)).subs(x,7.04e-3)
1.95377896947390

And yes, that matches what the voltmeter shows.

Let's try with \$2\:\text{k}\Omega\$ and get \$3.55\:\text{mA}\$ and find:

(6*x+1.75*.026*ln(1+x/4e-21)).subs(x,3.55e-3)
1.90168691368984

And yes again, that's what I got from Tinker for the LED voltage.

Let's up the current a bit so that \$R_\text{D}\$ matters more. Try with \$150\:\Omega\$ and get \$44.5\:\text{mA}\$:

(6*x+1.75*.026*ln(1+x/4e-21)).subs(x,44.5e-3)
2.26243555583850

And that's also the voltage I get from Tinker for the LED voltage.

Tinker uses the Shockley diode equation to solve these problems. So it doesn't need to guess. It uses a model and the model behaves in a certain way.

But Spice programs are really fast and good at bookkeeping and numerical solution solving, where we humans are rather poor and slow at both of those. So Spice programs use what they are good at.

We humans have to find other ways.

You've got a number of resonances by others on your question. But I'll dive directly in to demonstrate that it's not an assumption by Tinker.

The Shockley diode model can be simplified for these purposes to the following:

$$I_\text{D}=I_\text{SAT}\cdot\left[\exp\left(\frac{V_\text{D}-R_\text{D}\cdot I_\text{D}}{\eta\,V_T}\right)-1\right]$$

Solving that for \$I_\text{D}\$ would take a page's worth of math steps, which I'll avoid. (Uses branch-0 of LambertW function, which gives \$u\$ for \$y=u\, e^u\$ when you know \$y\$.)

But it can be rewritten in this more easily reckoned with equation:

$$V_\text{D}=I_\text{D}\cdot R_\text{D}+\eta\, V_T\cdot \ln\left(1+\frac{I_\text{D}}{I_\text{SAT}}\right)$$

Apparently, Tinker uses \$\eta=1.75\$, \$R_\text{D}=6\:\Omega\$, and \$I_\text{SAT}=4\times 10^{-21}\$ for its RED LED diode model. And \$V_T=26\:\text{mV}\$ for its thermal voltage.

In your case, with \$1\:\text{k}\Omega\$, you got \$7.04\:\text{mA}\$. So:

(6*x+1.75*.026*ln(1+x/4e-21)).subs(x,7.04e-3)
1.95377896947390

And yes, that matches what the voltmeter shows for the RED LED voltage.

Let's try with \$2\:\text{k}\Omega\$ and get \$3.55\:\text{mA}\$ and find:

(6*x+1.75*.026*ln(1+x/4e-21)).subs(x,3.55e-3)
1.90168691368984

And yes again, that's what I got from Tinker for the LED voltage.

Let's up the current a bit so that \$R_\text{D}\$ matters more. Try with \$150\:\Omega\$ and get \$44.5\:\text{mA}\$:

(6*x+1.75*.026*ln(1+x/4e-21)).subs(x,44.5e-3)
2.26243555583850

And that's also what I get from Tinker for the LED voltage.

Tinker uses the Shockley diode equation to solve these problems. So it doesn't need to guess. It uses a model and the model behaves in a certain way.

But Spice programs are really fast and good at bookkeeping and numerical solving methods, where we humans are rather poor and slow at both of those. So Spice programs use what they are good at.

We humans find other ways.