Yes, you can have any arbitrary, strictly positive constant instead of 1.
Why? First some background.
Math and separating hyperplane:
Support vector machines attempts to find a separating hyper-plane between sets $X$ and $Y$. Mathematically, the condition for a separating hyperplane is:
$$ \boldsymbol{w} \cdot \boldsymbol{x}_i - b < 0 \quad \quad \boldsymbol{w} \cdot \boldsymbol{y}_i - b > 0 $$
Observe that the inequalities are strict!
Numerical issues and practical solution:
Numerically, this formulation has practical problems. If the inequalities aren't strict, $\boldsymbol{w} = \boldsymbol{0}, b = 0$ is a trivial solution. Numerical optimization routines may give bizarre answers to this problem; standard floating point math isn't infinitely precise etc...
What to do? Let's replace the strict inequalities with non-strict inequalities plus some separation constant $t>0$:
$$ \boldsymbol{w} \cdot \boldsymbol{x}_i - b \leq -t \quad \quad \boldsymbol{w} \cdot \boldsymbol{y}_i - b \geq t $$
Yay! Numerical optimization can handle this. Also observe that since $\boldsymbol{w}$ and $b$ are choices variables, the scale of $t$ really doesn't matter. It's totally arbitrary. So we can just make it simple for ourselves and choose 1. (You could even choose different positive values for both inequalities; it doesn't matter.)
$$ \boldsymbol{w} \cdot \boldsymbol{x}_i - b \leq -1 \quad \quad \boldsymbol{w} \cdot \boldsymbol{y}_i - b \geq 1 $$
Other interpretation:
As your text explains, another interepretation of this is that you're fitting two parallel hyperplanes, one touching the X set, one touching the Y set, with some distance between them.