Try to complete your basis using the vectors of the standard basis: $(1,0,0,0), (0,1,0,0),$ etc. Pick one of them and check if the three vectors you have now are independent. If yes, great, keep the vector you chose and repeat the procedure with the three remaining vectors of the standard basis. If not, discard the vector you chose and try again with another one. Repeat until done.
The span of the vectors you mention is a subspace of the vector space to which they belong. So $(1,0,1,0)$ and $(0,1,0,1)$ span a two dimensional subspace (since as you note the vectors are linearly independent) of $\mathbb{R}^4$. It is not $\mathbb{R}^2$ exactly as this subspace contains $4$-tuples, but it is `equivalen' (in some sense) to $\mathbb{R}^2$ if this is of interest to you.
So in general a span of $m<n$ linearly independent vectors in $\mathbb{R}^n$ is a $m$-dimensional subspace of $\mathbb{R}^n$.
Best Answer
Using only basic analytic geometry: find three different non-collinear points on the plane, for example
$$A=(1,0,1)\;,\;\;B=(1,0,0)\;,\;\;C=(0,1,0)$$
and now construct the directed vectors
$$\vec{AB}=B-A=(0,0,-1)\;,\;\;\vec{AC}=C-A=(-1,1,-1)$$
and then the plane is
$$\pi:\;A+r\vec{AB}+s\vec{AC}=(1,0,1)+r(0,0,-1)+s(-1,1,-1)\;,\;\;r,t\in\Bbb R$$
Check: take the vectorial product of the direction vectors to get a perpendicular vector to the plane:
$$\vec{AB}\times\vec{AC}=\det\begin{vmatrix}i&j&k\\0&0&-1\\-1&1&-1\end{vmatrix}=(1,1,0)$$
and thus our plane is $\;x+y+d=0\;$ , and to find $\;d\;$ we can substitute any point on the plane here, say $\;A\;$ , to obtain
$$0+0+d=0\implies d=0$$
and the wanted plane is
$$x+y-1=0$$
which, of course, it is the same as you give. This way is just a standard form to check that what we got at the beginning is correct.
Anyway, you can look at your plane as the translation of a subspace, so;
$$\pi:\;\;\text{Span}\,\left\{\;(0,0,-1)\,,\,\,(-1,1,-1)\;\right\}+(1,0,1)$$