If I understand correctly you are asking two questions. The first is about an apparent issue because you are not using the fact that $y_0 = y_{-1} + x_0$. The second is if you can solve difference equations using the z-transform.
A few comments first:
My discussion ignores issues of convergence and existence of the z-transform. I assume that the relevant transforms are defined on $|z|>R$ for some $R$.
Given a sequence $x: \mathbb{Z} \to \mathbb{R}$, the unilateral z-transform of $x$ is given by $\hat{x}(z) = \sum_{n=0}^\infty x_n z^{-1}$.
A sequence is called causal iff $x_n = 0$ for all $n <0$.
Define the right-shift $(Rx)_n = x_{n-1}$. Note that $(Rx)_0 = x_{-1}$. Then we have $\widehat{Rx} (z) = x_{-1}+ z^{-1} \hat{x}(z)$.
If $x$ is causal (in fact, if $x_{-1} = 0$), then $(Rx)_0 = 0$ and $\widehat{Rx} (z) = z^{-1} \hat{x}(z)$.
If $x$ is causal, then it should be clear that $\widehat{R^kx} (z) = z^{-k} \hat{x}(z)$, where $R^k$ is $R\circ \cdots \circ R$ (that is, $k$ times).
Define the left-shift $(Lx)_n = x_{n+1}$. Note that if $x$ is causal , then $Lx$ is not necessarily causal because we have $(Lx)_{-1} = x_0$. In this case, we have $\widehat{Lx} (z) = z (\hat{x}(z)-x_0)$.
The point is that with causal sequences and the unilateral z-transform, a right shift corresponds to multiplication by $z^{-1}$, but a left shift needs to be modified slightly before multiplying by $z$.
Now for the first issue:
In the question above, the equation is $y_n = y_{n-1}+x_n$. Note that this can be written as $y_n = (Ry)_n + x_n$ or more succinctly as $y = Ry+x$.
Furthermore, setting $n=0$ gives $y_0 = y_{-1} + x_0$.
Taking the transform gives $\hat{y}(z) = y_{-1} + z^{-1} \hat{y}(z) + \hat{x}(z)$. This gives $\hat{y}(z) = {y_{-1}+\hat{x}(z) \over 1- z^{-1}}$.
If $y$ is causal, then $y_{-1} = 0$ and the formula simplifies to the formula above, that is, $\hat{y}(z) = {\hat{x}(z) \over 1- z^{-1}}$. So far, so good.
If I understand what it is that you did above (after the phrase "suggest that we"), you consider the equivalent equation $y_{n+1} = y_n + x_{n+1}$, which can be written as $Ly = y+Lx$. Taking transforms gives
$z ( \hat{y}(z)-y_0) = \hat{y}(z) + z ( \hat{x}(z)-x_0)$, dividing across by $z$ gives $\hat{y}(z)-y_0 = z^{-1} \hat{y}(z) + \hat{x}(z)-x_0$, which simplifies to $\hat{y}(z) = {y_{0}-x_0+\hat{x}(z) \over 1- z^{-1}}$.
Since $y_0 = y_{-1} + x_0$, the two equations are equal.
Now the second question:
If two different sequences $x,y$ satisfy $x_n = y_n $ for all $n \ge 0$ then we have $\hat{x} = \hat{y}$, that is, the unilateral z-transform only 'cares' about the values of $x$ when $n \ge 0$. Hence if we use this transform to solve difference equations, at best it will tell us about solution behaviour for $n \ge 0$.
If $x,y$ are causal and $\hat{x} = \hat{y}$, then we have $x=y$. This is what allows us to find a solution to a difference equation.
Suppose you have a difference equation of the form (taken from the notes you referred to)
$y_n = \sum_{k=0}^M b_k x_{n-k} - \sum_{k=0}^N a_k y_{n-k}$, and both $x,y$ are causal. Then this can be written as
$y =\sum_{k=0}^M b_k R^k x - \sum_{k=0}^N a_k R^k y$, and taking transforms yields
$\hat{y}(z) = \sum_{k=0}^M b_k z^{-k} \hat{x}(z) - \sum_{k=0}^N a_k z^{-k} \hat{y}(z)$, or equivalently,
$\hat{y}(z) = \hat{h}(z) \hat{x}(z)$, where
$\hat{h}(z) = { \sum_{k=0}^M b_k z^{-k} \over 1+\sum_{k=0}^N a_k z^{-k}}$.
Hence if one is given a suitable causal $x$, then to solve the equation one computes $\hat{x}$, multiplies it (pointwise) by $\hat{h}$ to get $\hat{y}$ and then computes the sequence that $y$ that corresponds to $\hat{y}$. Then this will be the unique causal solution to the difference equation.
