What you have highlighted is not true even for low speeds. If I move with a ball the trajectory doesn't exist, roughly speaking. But for other observers, it can be along any straight line if the ball doesn't accelerate.
If it accelerates, then (since you are talking about special relativity) I will no longer move with the ball. Suppose it were going in circles in my frame. It would of course, move in a helical path with respect to observer moving along a line perpendicular to plane of a circle. No, the shapes can be very crazy.
Secondly, you can construct a four vector in space time, which is by definition any quantity transforming like position vector of space time. The norm of this vector is a Lorentz scalar. Rather the inner product of any two four vectors is a Lorentz scalar. Mass being a scalar follows from this method easily.
For a particle with energy E and momentum $\mathbf{p}$, consider a vector $P = (E/c, p_x, p_y,p_z)$. This is a shorthand for representing components of a four dimensional vector. It can be shown that this transforms like position vector $X = (ct, x, y, z)$. Let $A = (a_0, \mathbf{a}), B = (b_0, \mathbf{b})$. The inner product of two four vectors is defined by (with respect to Lorentzian metric)
$$A\cdot B = a_0b_0 - \mathbf{a\cdot b}$$ where bold faced dot product is usual vector product in $\mathbb{R}^3$. When $A = B = P$, we get
$$ E^2/c^2 - \mathbf{p\cdot p} = P\cdot P$$
The right side is Lorentz scalar by claim, so left side,
$$ (m_0 c \gamma)^2 - (m_0v\gamma)^2 = m_0^2c^2$$
is too. In vacuum c is universal constant and thus result follows. Here $\gamma^{-2} = 1 - v^2/c^2$.
The space time interval invariance follows from norm of $X$. Using this method, sometimes you can end up with trivial scalars like speed of light, but sometimes you can also get differential equations which hold everywhere.
With four divergence operator vector $\nabla = (c^{-1}\partial_t, -\partial_x, -\partial_y, -\partial_z)$ and four-current ${J} = (c\rho, \mathbf{J})$ the equation $\nabla\cdot J = 0$ (sort of like a dot product) leads to equation of continuity in all frames. Here $\rho$ is mass/charge density and $\mathbf{J}$ is current density.
Note that while taking dot product space part goes with a minus sign to get correct equation.
