The answer is: 8 isomorphism classes. The classification up to absolute isomorphism yields: 7.

To show this in a greater and more natural generality, let in general $K$ be a field. I claim that there are 7 isomorphism classes of 2-dimensional $K$-algebras that are not (commutative) fields. Denote by $0_i$ the $i$-dimensional vector space endowed with null product. The isomorphism classes are represented by the algebras with basis $xy$ for which the nonzero products among $xx,xy,yx,yy$ are described

$W_1$: $xx=x$, $xy=y$;

$W_2$: opposite to $W_1$ ($xx=x$, $yx=y$);

(the remaining ones are commutative)

$W_3$: $0_2$;

$W_4$: $K\times 0_1$: $xx=x$;

$W_5$: $xx=y$;

(the remaining ones are unitary)

$W_6$: $xx=x$, $xy=yx=y$ ($K[t]/t^2$);

$W_7$: $xx=x$, $yy=y$ ($K\times K$).

This being granted (see below for a proof), it remains to classify quadratic extensions of $K$. If $K$ has characteristic $\neq 2$, they are classified by $K^*/K^{*2}\smallsetminus\{1\}$, which has order 2 when $K$ is finite. [This corresponds to twisted forms of $W_7$, if we include $W_7$ itself].

If $K$ has characteristic $2$, the separable quadratic extensions are classified by the cokernel of the additive endomorphism $t\mapsto t+t^2$ of $K$ (minus $\{0\}$), and again this cokernel has order 2, and again, including $W_7$, this are forms of $W_7$.

This covers finite fields. If we are interested in non-perfect fields of characteristic 2, we get the inseparable quadratic extensions, which are twisted forms of $W_6$. Including $W_6$, these are classified by the orbits of $K^{*2}\rtimes K^{*2}$ on $K$, acting by homotheties.

It remains to prove the classification above $W_{1\le i\le 7}$. Let $A$ be a 2-dimensional associative algebra.

First, the commutative unital case, which is the most standard (already covered by the previous answer): being artininan, if not local, $A$ is a product of at least 2 local unital commutative algebras, and the only possibility is $W_7$. If local and not a field, the maximal ideal $M$ has dimension 1, $M^2=0$ and we get $W_6$.

Second assume it commutative, but not unital. Adding a unit realizes our algebra $A$ as ideal of dimension $3$ in a 3-dimensional unital artinian algebra $A'$. If $A'$ is a product of fields, all its ideals are unital (as algebras, albeit not unital subalgebras). So the decomposition of $A'$ as product of local algebras involves a non-field. So either $A'$ is local, or $A'\simeq K\times K[t]/t^2$. In the second case, the two maximal ideals of $A'$ are $\{0\}\times K[t]/t^2$ (which is unital), and to $W_4$. Next suppose $A'$ is local. If $x^2=0$ for all $x\in A$, then in characteristic not $2$, we deduce $xy=0$ for all $x,y$ by polarization, and we have $A\simeq W_3$. If $x^2\neq 0$, then setting $y=x^2$ we see that $A\simeq W_5$.

(In char. 2 it remains to exclude another possibility, where $x^2=0$ for all $x$. If the product were nonzero, let $z$ be in its image, complete to a basis and rescale to get $xx=zz=0$, $xz=zx=z$, and then $x(xz)-(xx)z=xz-0=z$, contradicting associativity.)

Third and last, assume $A$ not commutative. There's a single non-commutative 2-dimensional Lie algebra up to isomorphism, namely with basis $(x,y)$ such that $[x,y]=y$. Hence (applying this to the associated Lie commutator) we can suppose that $A$ has a basis $(x,y)$ with $xy-yx=y$. Since $A$ is non-commutative and has dimension $\le 2$, the centralizer of every nonzero element is a proper subspace, and hence (using associativity), $x^2$ is a scalar multiple of $x$ for every $x$. So, we can write $xx=ax$, $yy=by$, $yx=cx+dy$, $xy=cx+(d+1)y$.

Then (u) $0=y(yx)-(yy)x=y(cx+dy)-(by)x=c(cx+dy)+dby-b(cx+dy)=c((c-b)x+dy)$.

Assume by contradiction $c\neq 0$. By (a) we deduce $d=c-b=0$. Thus $xx=ax$, $yy=cx$, $xy=cx+y$, $yx=cx$. Then $0=(xy)x-x(yx)=(cx+y)x-x(cx)=yx$, so $cx=0$, contradiction. Hence $c=0$, contradiction.

So $c=0$: $xx=ax$, $yy=by$, $yx=dy$, $xy=(d+1)y$.

Then $0=y(xy)-(yx)y=((d+1)-d)y^2=y^2=by$, so $b=0$.

Next $0=(yx)x-y(x^2)=dyx-y(ax)=(d-a)dy$. So $(d-a)d=0$. Also $0=x(xy)-(x^2)y=(d+1)xy-axy=(d+1-a)(d+1)y$. So $(d+1-a)(d+1)=0$.

If $d=0$, the latter writes $(1-a)=0$, so $a=1$, and $A\simeq W_1$. Otherwise $d=a$, and the latter writes $d+1=0$, so $d=-1=a$. In this case $A\simeq W_2$ (change $x$ into $-x$ to get the defining basis of $W_2$).

Finally the $W_i$ are pairwise non-isomorphic: we see that $W_1$ and $W_2$ are not isomorphic because $x$ is a left unit in $W_1$ while $W_2$ has no left unit. The other cases are straightforward.