$MGL$ does not admit a structure of $H\mathbb Z$-module. There are many ways to prove this. As Sean said in the comments, if it were true over $\mathbb C$, topological realization would imply that $MU$ is an $H\mathbb Z^{top}$-module, which is false. By rigidity this takes care of all characteristic zero fields. The same argument can presumably be made in positive characteristic using etale realization over a separable closure, but that would require computing the $l$-completed etale homotopy type of Thom spaces over Grassmannians which could be tricky.

Another argument is the following. If $MGL$ were an $H\mathbb Z$-algebra, $KGL$ would be too. Taking mapping spectra in $SH(k)$, this would imply that the $K$-theory spectrum $K(k)$ is an $H\mathbb Z^{top}$-module, which is false, as the first $k$-invariant of $K(k)$ is nontrivial (if the characteristic of $k$ is not $2$).

Here's an expansion of my comment about the $b_i$'s.
Let's use $x_i$ for the elements in the Hopkins-Morel isomorphism. These are homogeneous generators of the Lazard ring $L$, and there is no canonical choice for them. The Hurewicz map $L \to \mathbb Z[b_1,b_2,...]$ is injective but not surjective, though it becomes surjective after tensoring with $\mathbb Q$.
A complete formula for $x_i$ in terms of the $b_i$ depends of course on a specific choice of the generators $x_i$. You can see one such choice and the resulting formula in Hazewinkel Constructing formal groups II, equation (7.5.1), where $U_n$ is $x_{n-1}$ and $m_n(U)$ has degree $n-1$ and is determined by $\sum_{n\geq 1}m_n(U)(\sum_{i\geq 0}b_i)^n=1$.
Usually one only cares about the image of $x_i$ modulo decomposable, which is $\pm pb_i$ if $i+1$ is a power of a prime $p$ and $\pm b_i$ otherwise (with Hazewinkel's choice of generators the signs are minus signs).