Are these traits for my planet mathematically accurate? Could it support life?

Your star is not going to work, so the planet won't either.

Mass of star: 1.04 M☉

Radius of star: 0.69 R☉

Surface temperature of star: 4,620° K

Luminosity of star: 0.2 L☉

There is a well known set of relationships for mass-luminosity (and hence temperature) for stars. It's explained on this Wikpedia page.

For stars of around 1 solar mass that relationship says :

$$frac L {L_odot}= left( frac M {M_odot}right)^4$$

So your star with $M=1.04 M_odot$ should have $L=1.17L_odot$ and not $0.2L_odot$ as you state.

The knock on from this is of course the temperature and conditions on your planet.

The star is basically about 16% more luminous than our own, but your planet is about half the distance we are from the Sun. It's going to be hot !

The (very rough) calculation for effective temperature of a planet will be :

$$T = T_oplus left( frac {L(1-A)a_oplus^2} {L_odot(1-A_oplus)a^2} right)^{frac 1 4}$$

With your number for the star this gives $T approx 1.4 T_oplus$. So $40$ % hotter than earth, or about $110^circ K$ hotter. That puts it well into runaway greenhouse territory and it would likely be far worse.

Most of that is a result of your very close orbit - move it out past $1, AU$ and you'll get better results.

The star's radius is actually a bit tricky - at around $M=M_odot$ the typical mass-radius behavior has a distinct boundary - different behaviors on each side. Probably the simplest thing is to treat your star as being almost the same as the Sun in terms of mass and you get a radius of about the same as the Sun.

Stellar Heating

(using Stefan-Boltzmann equation) ${P over A} = epsilon sigma T^4$. Power per unit area at the surface of your star (0.69 suns ~ 480,000 km) is 2.58$times 10^7$ Watts per square-meter.

At perihelion, the planet is 72 million kilometers distant. $R2 over R1$ (where R1 is the solar surface radius) is 163, and the power received per unit area is

$P_1 times {R2 over R1}^2$ = 970.44 Watts per meter squared. At aphelion, 871.94 $W over m^2$.

Compared to Earth receiving around 1,374 $W over m^2$ peak, this planet gets about 63% to 70% to same amount of radiant heat.

When you include the lower Bond albedo, 619 to 689 $W over m^2$ peak make it to the surface.

Compared to Earth, this world holds on to 72% as much radiant heat.

$P_{average}$ = $P_{peak} over 4$. Therefore, the average power received is 163 $W over m^2$. The predicted surface temperature, then is 231 Kelvin / -41 Celsius; compared to Earth's predicted surface temperature of -18 C.

I think it should be pointed out that Earth's observed surface temperature is +14 C, and that Earth gets a +32 degree C boost from greenhouse effect.

This planet may support life, but there must be some sort of greenhouse or other heating effect raising the mean temperature above freezing.

Gravity

$M2 over M1$ = 2.57; $R2 over R1$ = 1.37; $G2 over G1$ = 1.36 gs. About what you calculated.