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This is my attempt I know this is incomplete or may even be wrong.

Let $θ_1 in arg(z)$ and $θ_2 in arg(w)$. Then, $θ_1+θ_2 in arg(z)+arg(w)$.

Also, $θ_1+θ_2 in arg(zw)$. Is this sufficient for the proof or correct at all? Hope someone could help me out. Thanks.

EDIT: Just to avoid any confusions I will add how $arg z$ is defined.

$arg z=textArg, z+2pi kmid ktext is an integer$

Let $z=r_1 e^i theta_1$ and $w=r_2 e^i theta_2$, then
$
zw=r_1 r_2 e^i (theta_1+theta_2).
$
Hence
$$
arg(z)=theta_1 + 2kpi, kin mathbfZ\
arg(w)=theta_2+2kpi, kin mathbfZ\
arg(zw)=theta_1+theta_2+2kpi, kin mathbfZ.
$$
There it is: $arg(zw)=arg(z)+arg(w)$.

Here's a proof: Let $z=r e^itheta$ and $w=s e^iphi$. Then $$textArg(zw)=textArgleft(rse^ithetae^iphiright)=textArgleft(rse^i(theta+phi)right)=textArg(z)+textArg(w)quad(textmod 2pi),$$
where $textArg(z)in[0,2pi)$ is the principal argument of $z$.

However, as Micah commented above, it looks like you're considering sets of all arguments, i.e. to obtain all representations, $$z=re^i(theta + 2pi n)$$ and $$w=se^i(phi+2pi m),$$ where $n,minmathbbZ$.

I'm not sure we can use the $w = e^itheta_w$ here because this is a notation that is partially induced from what he is trying to prove. Here's a proof that doesn't need this.

So we have $textarg(w) in theta_w$ and $textarg(z) in theta_z$.

This means $w = |w|(costheta_w+sintheta_w)$ and $z = |z|(costheta_z+sintheta_z)$.

By multiplication : $wz = |w| |z| ((cos theta_w cos theta_z - sin theta_w sin theta_z) + i (cos theta_w sin theta_z + sin theta_w cos theta_z))$ which happens to be equal to $|w| |z| (cos (theta_w + theta_z) + i sin (theta_w + theta_z)).$

Hence, noting $wz = x + iy$ and identifying the real and the imaginary part, we have $dfrac x = cos (theta_w + theta_z)$ and $dfrac y = sin (theta_w + theta_z)$.

This is the definition of $(theta_w + theta_z)$ belonging to $textarg(wz)$.