The column space is not a list of vectors so it's not clear what you mean when you ask if you should exclude non-pivot columns. The column space is the linear span of the columns. Each column (including the non-pivot columns) is contained in this space.
What you may be confusing yourself with is the column space vs. a basis for the column space. A basis is indeed a list of columns and for a reduced matrix such as the one you have a basis for the column space is given by taking exactly the pivot columns (as you have said). There are various notations for this, $\operatorname{Col}A$ is perfectly acceptable but don't be surprised if you see others.
As for the dimension of the column space, it's $3$, which is the number of elements in a basis, i.e., the number of pivot columns.
Lemma 1: Given an $m\times n$ matrix $A,$ the null space of $A^T$ is the orthogonal complement of the column space of $A.$
Proof: Write $A=[c_1\:\cdots\:c_n]$ where the $c_j$ are the columns of $A,$ and note that for any $m$-dimensional vector $x$ we have $$A^Tx=\left[\begin{array}{c}c_1^T\\\vdots\\c_n^T\end{array}\right]x=\left[\begin{array}{c}c_1^Tx\\\vdots\\c_n^Tx\end{array}\right]=\left[\begin{array}{c}c_1\cdot x\\\vdots\\c_n\cdot x\end{array}\right].$$ Since the column space of $A$ is spanned by $c_1,...,c_n$, then $x$ is in the orthogonal complement to the column space of $A$ if and only if $x$ is orthogonal to each $c_j$ if and only if each $c_j\cdot x=0$ if and only if $A^Tx$ is the $n$-dimensional zero vector if and only if $x$ is in the null-space of $A^T.$ $\Box$
Lemma 2: Let $V,W$ be subspaces of some finite-dimensional space $X$. $V$ and $W$ have the same orthogonal complement if and only if $V=W$.
Proof: If $V=W$, then their orthogonal complements are trivially the same.
Suppose $V,W$ have the same orthogonal complement. Take $v\in V$. Since $X$ is the direct sum of $W$ and its orthogonal complement, and since $x\in V\subseteq X$, then there exist unique $w,w'$ such that $v=w+w',$ $w\in W$, and $w'$ in the orthogonal complement of $W$. Since $V,W$ have the same orthogonal complement, then $w'$ is orthogonal to $v,$ and so $$0=v\cdot w'=(w+w')\cdot w'=w\cdot w'+w'\cdot w'.\tag{$\star$}$$ Since $w'$ is in the orthogonal complement of $W$ and $w\in W$, then $w\cdot w'=0$, so it follows by $(\star)$ that $$w'\cdot w'=0.$$ Now, no non-zero vector is self-orthogonal, so $w'$ must be the zero vector, whence $v=w\in W$, and so $V\subseteq W$. By symmetrical arguments, we likewise have $W\subseteq V$, so $V=W$. $\Box$
Proposition: Given matrices $A,B$ of the same dimensions, $A$ and $B$ have the same column space if and only if $A^T$ and $B^T$ have the same reduced row echelon form.
Proof: Let $rref(M)$ indicate the reduced row echelon form of a matrix $M$. Recall that we can obtain $rref(M)$ by Gauss-Jordan elimination, which involves multiplication on the left by some finite collection of elementary matrices--that is, for any $M$, there exist elementary matrices $E_1,\cdots,E_n$ of appropriate dimension such that $rref(M)=E_n\cdots E_1M.$ This collection of elementary matrices is not unique, but that isn't important. Note, though, that elementary matrices are invertible, so it follows that the null spaces of $rref(M)$ and $M$ are the same.
Thus, $A^T$ and $B^T$ have the same reduced row echelon form if and only if they have the same null space. By Lemma 1, $A^T$ and $B^T$ have the same null space if and only if the column spaces of $A$ and $B$ have the same orthogonal complement. By Lemma 2, the column spaces of $A$ and $B$ have the same orthogonal complement if and only if the column spaces of $A$ and $B$ are the same. $\Box$
Upshot: The Proposition lets us get around needing to know what the column spaces of two matrices are, and simply determine whether they have the same column space by converting their transposes to reduced row echelon form.
Best Answer
When you row-reduce a matrix, the dimension of the column space stays fixed, so if $A,B$ have the same reduced echolon form then the dimensions of the column spaces are equal, but the column spaces might not be equal: $$A=\begin{pmatrix}1&2\\1&2\end{pmatrix}\hspace{10pt}B=\begin{pmatrix}1&2\\2&4\end{pmatrix}$$ The have the same reduced echolon form, but different column-spaces.
In general, the only way to make sure that two matrices have the same column space is to column-reduce them (unless both are of full rank).