No. This group contains, e.g., $\begin{pmatrix}0&1\\-1&0\end{pmatrix}^2=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$.
Addendum: Let $A=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ and $B=\begin{pmatrix}0&i\\-i&0\end{pmatrix}$. We have $A^4=I, B^2=I, AB=BA$. Hence the group generated by them is $\langle A\rangle \times \langle B \rangle $ and has order 8.
Sorry, there are some issues here caused by your choice of notation. By multiplying the same $a,b,c,d$ matrix with both generators, you've overlooked that there's nothing wrong with multiplying the generators with two different matrices and adding, which will produce many more possiblities.
In the left ideal generated by those two things, a general element will look like this:
$\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}1&0\\0&0\end{bmatrix}+\begin{bmatrix}e&f\\g&h\end{bmatrix}\begin{bmatrix}0&1\\0&0\end{bmatrix}=\begin{bmatrix}a&e\\c&g\end{bmatrix}$.
Notice how we didn't recycle the $a,b,c,d$ matrix for both generators. Since you can pick $a,e,c,g$ to be whatever you want, you can see that the left ideal generated by these two things is the entire matrix ring.
Try to apply similar reasoning to the right ideal generated by these two things. You will get a different answer than the left ideal, and the answer you gave is not incorrect. It's just that you can express what the right ideal looks like more simply.
Best Answer
Here's a more structural approach. Let $G=\langle A,B\rangle$. As you already noted $A$ and $B$ have order $4$. Now note that $A^2 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$. Hence $P:=A^2B= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ is a permutation matrix. Clearly $G=\langle A,P\rangle$.
If $D$ is a diagonal matrix, then $PDP^{-1}$ is again a diagonal matrix with the diagonal entries flipped. Let $N$ be the subgroup of $G$ generated by $A$ and $PAP^{-1}$. Then $N$ consists of diagonal matrices with entries in $\{1,i,-1,-i\}$. Note that $N \cong \mathbb Z/4 \mathbb Z \times \mathbb Z/4\mathbb Z$. $N$ contains $\langle A \rangle$ and is closed under conjugation by $P$. Since $G$ is generated by $A$ and $P$, this is sufficient to see that $N$ is a normal subgroup of $G$. Moreover, $\langle P \rangle \cap N$ is the trivial group. Hence $G$ is a semidirect product of $N$ and $\langle P \rangle \cong \mathbb Z/2\mathbb Z$. Explicitly, $$ G \cong (\mathbb Z/4\mathbb Z \times \mathbb Z/4\mathbb Z) \rtimes \mathbb Z/2\mathbb Z, $$ where $\mathbb Z/2\mathbb Z$ acts on $\mathbb Z/4\mathbb Z \times \mathbb Z/4\mathbb Z$ by flipping the components.
In particular $G$ has 32 elements, and every element can be uniquely written in the form $A^j (PAP^{-1})^k P^l$ with $j$, $k \in \{0,\ldots,3\}$ and $l \in \{0,1\}$.