Let's define $X_t = \int_0^t \frac{\mathrm{d}B_s}{1-s}$, which is to say $\mathrm{d}X_t = \frac{1}{1-t} \mathrm{d}B_t$. Its sde thus has zero drift coefficient.
Then we are faced with using Ito formula for $W_t = (1-t) X_t$.
$$
\mathrm{d} W_t = \left(\frac{\partial ((1-t)X_t))}{\partial t} + \underbrace{0}_{\text{drift}} \frac{\partial ((1-t)X_t))}{\partial X_t} + \frac{1}{2} \left(\underbrace{\frac{1}{1-t}}_\text{diffusion}\right)^2 \underbrace{\frac{\partial^2 ((1-t)X_t))}{\partial X_t^2}}_0 \right) \mathrm{d}t + \frac{\partial ((1-t)X_t)}{\partial X_t} \mathrm{d}X_t
$$
Thus
$$
\mathrm{d}W_t = -X_t \mathrm{d}t + (1-t) \mathrm{d} X_t = -\frac{W_t}{1-t} \mathrm{d}t + \mathrm{d} B_t
$$
I think the given representation of the Brownian Bridge is not correct. It should read
$$Y_t = a \cdot (1-t) + b \cdot t + (1-t) \cdot \underbrace{\int_0^t \frac{1}{1-s} \, dB_s}_{=:I_t} \tag{1}$$
instead. Moreover, the covariance is defined as $\mathbb{E}((Y_t-\mathbb{E}Y_t) \cdot (Y_s-\mathbb{E}Y_s))$, so you forgot to subtract the expectation value of $Y$ (note that $\mathbb{E}Y_t \not= 0$).
Here is a proof using the representation given in $(1)$:
$$\begin{align} \mathbb{E}Y_t &= a \cdot (1-t) + b \cdot t \\
\Rightarrow \text{cov}(Y_s,Y_t) &= \mathbb{E}((Y_t-\mathbb{E}Y_t) \cdot (Y_s-\mathbb{E}Y_s)) = (1-t) \cdot (1-s) \cdot \mathbb{E}(I_t \cdot I_s) \\ &= (1-t) \cdot (1-s) \cdot \underbrace{\mathbb{E}((I_t-I_s) \cdot I_s)}_{\mathbb{E}(I_t-I_s) \cdot \mathbb{E}I_s = 0} + (1-t) \cdot (1-s) \mathbb{E}(I_s^2) \tag{2} \end{align}$$
for $s \leq t$ where we used the independence of $I_t-I_s$ and $I_s$. By Itô's isometry, we obtain
$$\mathbb{E}(I_s^2) = \int_0^s \frac{1}{(1-r)^2} \, dr = \frac{1}{1-s} -1.$$
Thus we conclude from $(2)$:
$$\text{cov}(Y_t,Y_s) = (1-t) \cdot (1-s) \cdot \left( \frac{1}{1-s}-1 \right) = s-t \cdot s = s \cdot (1-t).$$
Best Answer
If $f$ is any continuous function, then L'Hopital's rule gives \[ \lim_{t\to1}\; (1-t)\int_0^t\frac{f(s)}{(1-s)^2}\,ds = f(1). \] In terms of generalized functions, this says that if \[ \mu_t(s) = \frac{1-t}{(1-s)^2}1_{[0,t]}(s), \] then $\mu_t\to\delta_1$. The Brownian bridge can also be written in terms of generalized functions: \[ Y_t= a(1 - t) + bt + \langle\partial B,\nu_t\rangle, \] where \[ \nu_t(s) = \frac{1-t}{1-s}1_{[0,t]}(s), \] and $\partial B$ is the (random) distributional derivative of the Brownian sample path. Note that $\langle\partial B,\nu_t\rangle = -\langle\partial\nu_t,B\rangle$, and \[ \partial\nu_t = (1-t)\delta_0 + \mu_t - \delta_t. \] Hence, \begin{align*} \lim_{t\to1}\;\langle\partial B,\nu_t\rangle &= \lim_{t\to1}\;-\langle(1-t)\delta_0 + \mu_t - \delta_t,B\rangle\\ &= \lim_{t\to1}\;\langle\delta_t - \mu_t,B\rangle\\ &= \langle\delta_1-\delta_1,B\rangle = 0. \end{align*}