I think the last edit is correct.
But for the row operation part I would do
R1 = R1 + R3, R2 = R2 + R3
and then swap R1 and R3 to get an [I3|A]
Once you have $H$, to find $G$ you need to find a set of $k$ rows of length $n$ that are LI and orthogonal to the rows of $H$.
In a simple case like this, it could be done by trying, eg:
$$ G=
\begin{bmatrix}
1 & 1 & 0 & 0 & 0 \\
0 & 1 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 1 \\
\end{bmatrix}$$
A more general and systematic way is to try to find an equivalent code (same codebook, different mapping) by doing elementary operations with the rows of $H$, to put it in systematic form: $H=( I | P´)$ and then $G=( P | I)$ fits the bill.
In this case, we cannot do that manipulating the rows alone. We can resort to an aditional trick: you are also allowed to permute some columns , to bring it to systematic form, but then at the end un-permute the rows in the resulting $G$.
So, let's permute columns 2 and 4 to get the modified parity matrix
$$\begin{bmatrix}
1 & 0 & 1 & 1 & 0 \\
0 & 1 & 0 & 0 & 1 \\
\end{bmatrix}=[I | P']
$$
and the modified generator matrix is
$$[P | I] =\begin{bmatrix}
1 & 0 & 1 & 0 & 0 \\
1 & 0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 & 1 \\
\end{bmatrix}
$$
and after permuting columns 2 and 4 we get
$$ G = \begin{bmatrix}
1 & 0 & 1 & 0 & 0 \\
1 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 1
\end{bmatrix}
$$
You can (you should) check that the rows are indeed LI and orthogonal to the rows of $H$.
Best Answer
Hint: If you work over the field ${\Bbb F}_5$, then multiply the first row of $H$ with the inverse of $2$ and the second row with the inverse of $4$. Then exchange the rows of $H$. This gives you the desired form $H = (A\mid I)$ from which you can easily obtain the corresponding generator matrix.
More concretely, in $\left(\begin{array}{ccccc} 4&1&4&2&0&4\\2&1&1&4&2&0 \end{array}\right)$ multiply the first row by $4$ and the second row by $3$ giving $\left(\begin{array}{ccccc} 1&4&1&3&0&1\\1&3&3&2&1&0 \end{array}\right)$. Swapping rows gives $\left(\begin{array}{ccccc} 1&3&3&2&1&0\\1&4&1&3&0&1 \end{array}\right)$ as required.