Setting: Let $M$ be a smooth manifold and $\{U_\alpha\}_{\alpha \in \mathcal{I}}$ a locally finite covering of open subsets. Furthermore let $G$ be a smooth Lie group. Now assume we are given a family of transition functions $$\phi_{\alpha\beta}\in \mathcal{C}^0(U_\alpha \cap U_\beta; G) \quad\text{for any}\quad \alpha,\beta \in \mathcal{I}$$ satisfying the cocyclicity conditions: $$ \phi_{\alpha\alpha} = \mathbb{1}_{U_\alpha} \\ \phi_{\alpha\beta} \phi_{\beta\gamma} = \phi_{\alpha\gamma}. $$ on the obvious domains. Now $P := \left( \coprod_{\alpha \in \mathcal{I}} U_\alpha \times G \right)\big/{\sim}$ with $$(x,g)_\alpha \sim (x, \phi_{\beta\alpha}(x)\cdot g)_\beta \quad\text{for any}\quad x \in U_\alpha \cap U_\beta$$ defines a $G$-principal fibre bundle over $M$ (in the category of topological spaces).
Question: Can we conclude that $P \rightarrow M$ is a smooth $G$-principal bundle over $M$? That is, a principal bundle in the category of smooth manifolds?
Proof of continuous principal bundle is equivalent to smooth principal bundle
So the only thing left to do is to prove that your construction gives out a continuous principal bundle.
The first condition of been a continuous principal bundle is: The right $G$ action is continuous. Which in this construction, $(U_\alpha\times G)\times G\to(U_\alpha\times G)$, $(x,g)h\mapsto (x,gh)$, Because of the definition of Lie groups, it must be smooth, thus continuous.
And the second condition says that base manifold $M$ must exist local trivial cover $U_\alpha$, and that is already given in your construction.
So we conclude the $G$-principle bundle given is a continuous $G$-principal bundle. $\square$