I am reading a little about cobordism and I have a basic question, which makes sense both in the topological and motivic setting. Let $\mathrm{Gr}_{n,\infty}$ denote the infinite $n$-Grassmanian and denote $\xi_n $ its universal bundle. The cobordism spectrum $\mathrm{MGL}$ is made out of the universal Thom spaces.That is to say, it is the spectrum $$ (S\, ,\, \mathrm{Th}(\xi_1)\, ,\, \mathrm{Th}(\xi_2)\, ,\, \ldots \, ,\, \mathrm{Th}(\xi_i)\, ,\, \ldots \ ) $$ with the natural bounding maps. My question is

Is $\mathrm{MGL}[2r](r)$ isomorphic, in $\mathbf{SH}$, to the spectrum $(\, \mathrm{Th}(\xi_r)\, ,\, \mathrm{Th}(\xi_{r+1})\, ,\, \ldots \, ,\, \mathrm{Th}(\xi_{r+i})\, ,\, \ldots \ )$? And if so, why?

It seems to me that this should be right and simple, but since $\mathrm{MGL}$ is not an $\Omega$-spectrum I fail to see a direct reason.