Let $y \in \mathbb R$ and define $f_y\colon [0,1] \to \mathbb R$ by
$$ f_y(x) := \sum_{r=0}^\infty b_r(x)\frac{y^r}{r!} $$
We have, taking derivatives, that for $x \in [0,1]$:
\begin{align*}
f_y'(x) &= \sum_{r=0}^\infty b_r'(x) \frac{y^r}{r!}\\
&= \sum_{r=1}^\infty rb_{r-1}(x) \frac{y^r}{r!}\\
&= y\cdot \sum_{r=0}^\infty b_r(x) \frac{y^r}{r!}\\
&= y \cdot f_y(x)
\end{align*}
Hence $f_y(x) = \exp(xy)f_y(0)$. Integrating, we have by uniform convergence
\begin{align*}
\int_0^1 f_y(x)\, dx &= \sum_{k=0}^\infty \int_0^1 b_r(x)\, dx \cdot \frac{y^r}{r!}\\
&= 1.
\end{align*}
On the other hand
\begin{align*}
\int_0^1 f_y(x)\, dx &= \int_0^1 f_y(0)\exp(xy)\, dx\\
&= f_y(0) \cdot \left.\frac{\exp(xy)}y\right|_{x=0}^1\\
&= f_y(0) \cdot \frac{\exp y -1}y
\end{align*}
So
$$ 1 = f_y(0) \cdot \frac{\exp y - 1}y \iff f_y(0) = \frac y{\exp y -1} $$
This gives
$$ f_y(x) = \frac{y\exp(xy)}{\exp y- 1}. $$