Proof that Quaternion Algebras are simple

I'll retract this if whatever $A=\left(\frac{a,b}{F}\right)$ means prohibits my example :)

I decided to run fast with my "what if $F$ has characteristic two" comment fast and mention why the quaternions over a field of characteristic 2 are never a simple ring.

Since $-1=1$ in such a ring, the generators wind up commuting instead of anticommuting, so the whole thing winds up commutative. Thus the ring would be simple iff it is a field. However, as mentioned, $(1+i+j+k)\cdot(1-i-j-k)=0$ in that characteristic, and so it is clearly not a field.

Thanks for clarifying the original post. That construction is just the Clifford algebra of a two dimensional space with the symmetric bilinear form $\begin{bmatrix}a&0\\0&b\end{bmatrix}$. You will find a lot said about this when the form is nondegenerate (that is, $a,b$ are both nonzero.) If you check out the link, you can see that for real algebras, an even dimensional vector space with a nondegenerate form is isomorphic to a full matrix ring over a division ring (which is necessarily simple).

I wish I could instantly recall if the same is true for rings of characteristic other than 2. My memory is telling me something like that appears (or at least, the proof for "quaternion algebras") in Jacobson's Basic Algebra I in the section on the Hurwitz problem (7.6, I see from the googlebooks preview).

The algebra $A=\left(\dfrac{a,b}{F}\right)$ is simple when $ab\neq0$, and $2\neq0$ in $F$.

You haven't used the fact that an ideal must be 2-sided. Let's see how that helps.

Let $I$ be any non-zero ideal, and let $q=x_0+x_1i+x_2j+x_3k$ be an arbitrary element non-zero element of $I$. Then using the multiplication rules (and their consequence $ki=-jii=-aj, ik=iij=aj$) we see that both $$ qi=-ax_1+x_0i-ax_3j-x_2k $$ and $$ iq=-ax_1+x_0i+ax_3j+x_2k $$ are in $I$. As $2$ is invertible in $F$ we have that both $$ \frac{iq-qi}2=ax_3j+x_2k\in I $$ and $$ \frac{iq+qi}2=-ax_1+x_0i\in I. $$ Furthermore, one of them is non-zero. Because $j$ is invertible $ax_3j+x_2k=(ax_3+x_2i)j$ and $ax_3+x_2i$ generate the same ideal, so we can deduce in either case that the ideal $I$ contains a non-zero element of the form $z=y_0+y_1i$. Then also the following elements are in $I$: $$ zj=y_0j+y_1k\in I\qquad\text{and}\qquad jz=y_0j-y_1k\in I. $$ Repeating the above trick this implies that $$ \frac{zj-jz}2=y_1k\in I\qquad\text{and}\qquad\frac{zj+jz}2=y_0j\in I. $$ At least one these non-zero. But any non-zero $F$-multiple of any of the basis elements is a unit in $A$, so we can conclude that $I$ contains a unit of $A$ and is thus all of $A$.

Compare this with the exercise you have probably had earlier. The algebra of 2x2-matrices over any field has no non-zero ideals (i.e. it is simple - the same applies to matrix algebras of all sizes). That algebra does have non-trivial left (resp. right) ideals. Consider the sets of matrices where a fixed column (resp. row) is constrained to be all zeros. The upshot here is that in non-commutative algebra 2-sidedness of an ideal is a much stronger requirement. The early examples of ideals a student sees are more often than not in commutative rings, and things are simpler there.