Sifting Property of Shifted Impulse
It would be better to show that property using pencil and paper, because you're fooling yourself with that Matlab code. Of course we have
$$\delta[n-n_0]\star g[n]=g[n-n_0]\tag{1}$$
Now define $h[n]=g[n-n_1]$. According to $(1)$ we have
$$\delta[n-n_0]\star h[n]=h[n-n_0]\tag{2}$$
And, consequently, with $h[n]=g[n-n_1]$,
$$\delta[n-n_0]\star g[n-n_1]=g[n-n_0-n_1]\tag{3}$$
No need to waste computing power to show this.