You want to prove that the sequence of partial sums $ \begin{eqnarray} S_N(x) = \sum_{n=0}^{N} a_n(x) b_n(x) \end{eqnarray} $ converges uniformly as $N \rightarrow \infty$. Observe that $ \begin{eqnarray} S_N(x) &=& \sum_{n=0}^{N} a_n(x) b_n(x) = \sum_{n=0}^{N} \left( \sum_{k=0}^n a_k(x) - \sum_{k=0}^{n-1} a_k(x) \right) b_n(x) \\ &=& \sum_{n=0}^{N} \left( \sum_{k=0}^n a_k(x) \right) b_n(x) - \sum_{n=0}^{N} \left( \sum_{k=0}^{n-1} a_k(x) \right) b_n(x) \\ &=& \sum_{n=0}^{N} \left( \sum_{k=0}^n a_k(x) \right) b_n(x) - \sum_{n=0}^{N+1} \left( \sum_{k=0}^{n-1} a_k(x) \right) b_n(x) + \left( \sum_{k=0}^{N} a_k(x) \right) b_{N+1}(x) \\ &=& \sum_{n=0}^{N} \left( \sum_{k=0}^n a_k(x) \right) b_n(x) - \sum_{n=0}^{N} \left( \sum_{k=0}^{n} a_k(x) \right) b_{n+1}(x) + \left( \sum_{k=0}^{N} a_k(x) \right) b_{N+1}(x) \\ &=& \sum_{n=0}^{N} \left( \sum_{k=0}^n a_k(x) \right) b_n(x) - \sum_{n=0}^{N} \left( \sum_{k=0}^{n} a_k(x) \right) b_{n+1}(x) + \left( \sum_{k=0}^{N} a_k(x) \right) b_{N+1}(x) \\ &=& \sum_{n=0}^{N} \left( \sum_{k=0}^n a_k(x) \right) (b_n(x) - b_{n+1}(x)) + \left( \sum_{k=0}^{N} a_k(x) \right) b_{N+1}(x). \end{eqnarray} $ (This manipulation is called partial summation or summation by parts.) So it will be enough to show that both $\sum_{n=0}^{N} \left( \sum_{k=0}^n a_k(x) \right) (b_n(x) - b_{n+1}(x))$ and $\left( \sum_{k=0}^{N} a_k(x) \right) b_{N+1}(x)$ converge uniformly as $N \rightarrow \infty$
Since $\left( \sum_{k=0}^{N} a_k(x) \right)$ is bounded and since $b_{n}(x)$ converges to $0$ uniformly, it follows that $\left( \sum_{k=0}^{N} a_k(x) \right) b_{N+1}(x)$ converges to $0$ uniformly.
I leave it to you to show that the boundedness of $\left( \sum_{k=0}^{n} a_k(x) \right)$ and the uniform convergence of $\sum_{n=0}^{N} |b_n(x) - b_{n+1}(x)|$ implies that $\sum_{n=0}^{N} \left( \sum_{k=0}^n a_k(x) \right) (b_n(x) - b_{n+1}(x))$ converges uniformly.
Remark: Instead of showing that $\sum_{n=0}^{N} |b_n(x) - b_{n+1}(x)|$ converges uniformly using the Cauchy Criterion and then that the boundedness of $\left( \sum_{k=0}^{n} a_k(x) \right)$ and the uniform convergence of $\sum_{n=0}^{N} |b_n(x) - b_{n+1}(x)|$ implies that $\sum_{n=0}^{N} \left( \sum_{k=0}^n a_k(x) \right) (b_n(x) - b_{n+1}(x))$ converges uniformly, I suggest just showing directly that $\sum_{n=0}^{N} \left( \sum_{k=0}^n a_k(x) \right) (b_n(x) - b_{n+1}(x))$ converges uniformly using the Cauchy Criterion. The argument is very similar to the one used to prove that $\sum_{n=0}^{N} |b_n(x) - b_{n+1}(x)|$ converges uniformly. Just add in the fact that $\left| \sum_{k=0}^{n} a_k(x) \right| \leq M$.