[guided]The purpose of this step is to turn local microlocal information into a global identity. We have finitely many elliptic operators $A_{j,h}$, and the ellipticity of $A_{j,h}$ on $U_{q_j}$ means that its principal symbol $a_j$ is invertible there. Because the open sets $U_{q_j}$ cover the whole compactified cotangent bundle $\overline{T}^*X$, we can choose symbols $b_j$ supported microlocally inside $U_{q_j}$ so that the products $b_j a_j$ add up to $1$ microlocally.
More precisely, let $a_j\in S^0(\overline{T}^*X)$ be the compactified principal symbol of $A_{j,h}$, where $S^0(\overline{T}^*X)$ denotes the order-zero compactified semiclassical symbol class. On $U_{q_j}$, the symbol $a_j$ is elliptic, so one may divide by $a_j$ there at the symbolic level. Choosing a smooth partition of unity on the compact space $\overline{T}^*X$ subordinate to the finite cover $U_{q_1},\dots,U_{q_J}$ gives a principal-symbol inverse. To upgrade this principal inverse to an $O(h^\infty)$ inverse, one uses the full semiclassical composition formula recursively: after the principal term has killed the order-one error, the next coefficient in the full symbol of each $b_j$ is chosen to kill the next power of $h$, and continuing this construction gives a full symbol whose remaining error lies in $h^\infty S^{-\infty}$. Here $S^{-\infty}$ is the residual compactified semiclassical symbol class, and $h^\infty S^{-\infty}$ means that every residual symbol seminorm is $O(h^N)$ for every $N\in\mathbb{N}$.
Now quantize each full symbol $b_j$ to an order-zero semiclassical pseudodifferential operator $B_{j,h}:C^\infty(X)\to C^\infty(X)$. The semiclassical composition calculus says that the full symbol of $\sum_{j=1}^J B_{j,h}A_{j,h}$ equals the full symbol of the identity modulo $h^\infty S^{-\infty}$. Therefore the sum of the compositions recovers the identity modulo a residual smoothing operator:
\begin{align*}
I = \sum_{j=1}^J B_{j,h}A_{j,h} + R_h.
\end{align*}
Here $R_h:\mathcal{D}'(X)\to C^\infty(X)$ is residual, meaning its Schwartz kernel is smooth and all of its symbol seminorms are $O(h^\infty)$. This is the point where compactness is essential: without a finite cover, we would not get a finite operator identity of this form.[/guided]