The answer is yes, you can!
The question of whether a planet can retain an atmosphere depends on the ratio between the escape velocity and the average thermal speed of particles in the atmosphere. Very, very roughly, if $v_{\mathrm{esc}}/v_{\mathrm{therm}}\gtrsim6$, the planet should be able to retain its atmosphere. A quick calculation shows that for nitrogen molecules, the ratio is a bit over 22, consistent with the fact that Earth's atmosphere contains quite a lot of nitrogen. For molecular hydrogen, on the other hand, the ratio is under 6 (though not by much), which explains why Earth has lost the hydrogen envelope it may have initially formed.
As your planet has an escape velocity very similar to Earth's and a temperature the same as Earth's, we would expect an atmosphere with a composition quite similar to Earth's - with heavier gases like nitrogen and oxygen, but little hydrogen and helium. (Earth actually isn't exceedingly far from the threshold for being able to retain some amount of helium, but you do have to account for factors like the solar wind's effect on the upper atmosphere.)
It's certainly possible to attain whatever pressure you want; simply increase the mass of the atmosphere! It turns out that we can relate the mass of the atmosphere $M_{\text{atm}}$ to the surface pressure $p_0$ by
$$M_{\text{atm}}=\frac{p_0}{g}4\pi R^2=\frac{4\pi R^4p_0}{GM}$$
with $M$ the mass of the planet and $R$ its radius. Given your values for escape velocity and surface gravity, we can calculate that it should have a radius 1.14 times that of Earth and a mass 1.16 times that of Earth$^{\dagger}$. We plug these values into the above equation, along with a pressure of 1.5 atm, and we find that your planet would need to have an atmosphere 2.26 times that of Earth. This holds regardless of atmospheric composition!
There's still the question of how realistic it is for a planet to accrete that much gas in the first place, and I don't have a good answer for that. It's possible that the conditions in the protoplanetary disk included higher concentrations of these heavier gases than you'd normally find.
$^{\dagger}$Since
$$v_{\mathrm{esc}}=\sqrt{\frac{2GM}{R}},\quad g=\frac{GM}{R^2}$$
we can show that $v_{\mathrm{esc}}=\sqrt{2gR}$, allowing us to calculate $R$ by
$$R=\frac{v_{\mathrm{esc}}^2}{2g}$$ once we have that, we can find $M$ by rearranging our equation for escape velocity:
$$M=\frac{v_{\mathrm{esc}}^2R}{2G}$$
