If "inside the modulus is alway positive", that is, if x+ 1> 0, then |x+ 1|= x+ 1 and so your equation becomes y= x+ 1- 1= x. The graph of y= x is the straight line passing through (0, 0).
If, however, you only meant that the modulus (absolute value) is never negative (it can be 0), then you need to break the problem into two parts: If x< -1, then x+1< 0 and so |x+ 1|= -(x+1). For x< -1, y= |x+1|- 1= -x- 1- 1= -x- 2. That crosses the x-axis when -x- 2= 0 or when x= -2 which is less that -1 so one x-intercept is at x= -2. The "y- intercept" occurs when x= 0 which does not satisfy x< -1.
If x\ge -1 then x+ 1\ge 0 and so |x+ 1|= x+ 1. NOW we have y= |x+1|- 1= x+ 1=1 = x which crosses the x-axis and y-axis at (0,0).
The x-intercepts are x= -2 and x= 0 and the y-intercept is y= 0.
