In a unital ring $R$ of order $p^2$ the additive order of $1$ can only be $p$ or $p^2$. In the second case, $1$ generates the additive group, so it is clear that the ring itself is isomorphic to $\mathbb Z/(p^2)$. So we concentrate in the first option.
In that case all non-zero elements have additive order $p$, so the ring is in fact a $2$-dimensional $\mathbb F_p$-algebra. Pick any element $x\in R$ which is linearly independent with $1$, so that $\{1,x\}$ is a basis. We must have $x^2=a+bx$ for some $a$, $b\in\mathbb F_p$, and then $R$ is isomorphic to $\mathbb F_p[X]/(X^2-aX-b)$.
If $X^2-aX-b$ is irreducible over $\mathbb F_p$, then $R$ is a field, and there is only one field of order $p^2$: $\mathbb F_{p^2}$. If not, then it factors as $(X-\alpha)(X-\beta)$ over $\mathbb F_p$. It is easy to see now that if $\alpha\neq\beta$ we have $R\cong\mathbb F_p\times\mathbb F_p$ and if $\alpha=\beta$ then $R\cong\mathbb F_p[X]/(X^2)$.
Interestingly, the fact that $R$ is commutative plays no role here, and there are no non-commutative rings of order $p^2$.
For fun, let us suppose now that $R$ does not have a unit, and let us see what happens.
First, suppose the addititive group is cyclic, so that there is an additive generator $x\in R$ of order $p^2$. Then there is an $n\in\{0,\dots,p^2-1\}$ such that $x^2=nx$. It is clear that the isoclass of $R$ is determined by $n$.
But there is an ambiguity, as there are other generators: any other generator of the additive group is of the form $ax$ with $a$ a unit of $\mathbb Z/(p^2)$.
If instead of $x$ we had started with $y=ax$, then as $y^2=a^2x^2=a^2nx=any$, instead of $n$ we would have found $an$. It follows that the isomorphism classes of rings of this form are parametrized by the quotient of $\mathbb Z/(p^2)$ under the action of its group of units given by left multiplication.
If I did not get this too wrong, there are three orbits: the one for $0$, the one for $1$ and the one for $p$, so there are three rings of this type.
Second, let us suppose that the additive group is not cyclic, so that all non-zero elements have order $p$, and $R$ is in fact a $\mathbb F_p$-vector space.
Suppose there is a non-zero idempotent $e\in R$. Let $\lambda:a\in R\mapsto ea\in R$ and $\rho:a\in R\mapsto ae\in R$. These are two idempotent $\mathbb F_p$-linear maps which commute. Linear algebra tells us then that there is a direct sum decomposition $$R=R_{00}\oplus R_{10}\oplus R_{01}\oplus R_{11}$$
with
\begin{align}
&R_{00}=\{x\in R:ex=xe=0\},\\
&R_{10}=\{x\in R:ex=x, xe=0\},\\
&R_{01}=\{x\in R:ex=0, xe=x\}, \\
&R_{11}=\{x\in R:ex=xe=x\}.
\end{align}
At most two of these subspaces can be non-zero, and we know that $R_{11}$ contains $e$. Also, $R_{11}$ cannot be all of $R$ because we are supposing there is no unit.
If there is a non-zero element in $R_{10}$, call it $x$. Then $\{e,x\}$ is a basis, $ex=x$, $xe=0$ and $xx=x(ex)=(xe)x=0x=0$. The multiplication is therefore completely determined.
Similarly, if there is a non-zero element $x\in R_{01}$, $\{e,x\}$ is again a basis, and we have $ex=0$, $xe=x$ and $xx=0$.
Finally, if there is a non-zero element $x\in R_{00}$, we have $ex=xe=0$. Then $ex^2=(ex)x=0$ and similarly $x^2e=0$, so $x^2$ is too in $R_{00}$, and there is a scalar $a$ such that $x^2=ax$. If $a\neq0$, we let $y=a^{-1}x$, so that $y^2=a^{-2}x^2=a^{-1}x=y$, and since $\{e,y\}$ is a basis this determines the multiplication. If $a=0$, of course the multiplication is also fixed.
We are left with the case in which there are no non-zero idempotents in $R$. A classical theorem of Albert implies that all non-zero elements must be nilpotent. If there is an $x\in R$ which is non-zero and such that $x^2$ is linearly independent with $x$, then $\{x,x^2\}$ is a basis, and we must have $x^3=0$, for the map $a\in R\mapsto xa\in R$, being a nilpotent endomorphism of a vector space of dimension $2$, must have nilpotency index at most $2$. We see that the multiplication table is completely determined here too.
Finally, suppose all non-zero elements are nilpotent but for for each $x$ of them, we have that $x^2$ is a linear multiple of $x$. All elements of $R$ must therefore square to zero. Let $\{x,y\}$ be a basis, and suppose $xy=ax+by$ and $yx=cx+dy$. Then $0=(x+y)^2=(a+c)x+(b+d)y$, so $yx=-xy$. Now call $u=xy$. If $u$ is zero, then all products are zero. If $\{x,u\}$ is a basis, then we know its the multiplication, as $x^2=u^2=xu=ux=0$. If not, $\{y,u\}$ is a basis, and again all products are zero.
Best Answer
The hard part of this is showing that the only commutative rings with additive group $(\mathbb{Z}/2\mathbb{Z})^3$ are the six you mentioned; the other additive groups have at most two generators and I imagine you've completed that part.
Here I'll give two proofs. The second proof is much shorter than the first.
First Proof
To this end, I will call the generators of $(\mathbb{Z}/2\mathbb{Z})^3$ $1,e_2$ and $e_3$ (WLOG the multiplicative identity is a generator, since any nonzero element can be a generator). And before fixing our perspective on $e_2$ and $e_3,$ I will split it into two cases: rings with a subring with additive group $(\mathbb{Z}/2\mathbb{Z})^2,$ and rings with no such thing.
Let's do the second case first. Notice that there are only three multiplications to fix to determine the entire ring: $e_2^2, e_2e_3,$ and $e_3^2.$ Now since we said that there is no subring with additive group $(\mathbb{Z}/2\mathbb{Z})^2,$ it must be true that $1, e_2,$ and $e_2^2$ generate the whole group; thus WLOG $e_3=e_2^2.$
Since now $e_3^2=e_2(e_2e_3),$ there is in fact only one multiplication to set, $e_2e_3.$ Of the eight possibilities, four immediately give errors: $e_2e_3$ cannot be equal to $0, e_2, e_3,$ or $e_2+e_3+1,$ or we would obtain that $\{0,1,e_3,e_3+1\}$ is a subring. Thus we are left with:
But in fact the first and fourth of these give errors as well, since $\{0,1,e_2+e_3,e_2+e_3+1\}$ is a subring in these cases. Another way to say the second and third of these is:
Both of these clearly define $\mathbb{F}_8.$
Now we consider the case where there is a subring isomorphic to $\mathbb{F}_4, \mathbb{F}_2[x]/(x^2),$ or $\mathbb{F}_2^2.$ In each case let us say that that subring is $\{0,1,x,x+1\},$ and the other four elements are $\{y,y+1,x+y,x+y+1\}.$ Either $xy$ is to be found in the former subring or it isn't. In the first case by choosing $y$ appropriately we may have $xy=0$ in the case of $\mathbb{F}_4,$ and we may have either $xy=0$ or $xy=1$ in the cases of $\mathbb{F}_2[x]/(x^2)$ and $\mathbb{F}_2^2.$ Since the value of $y^2$ hasn't been set yet, this leaves us a priori with 40 possibilities. But in case $xy=0,$ we must have that $y^2$ is annihilated by $x,$ which is 2/8 for $\mathbb{F}_4$ and 4/8 for $\mathbb{F}_2[x]/(x^2), \mathbb{F}_2^2.$ And in case $xy=1,$ we must have $xy^2=y,$ which has been rendered impossible. So there are only 10 possibilities, in each of which $xy=0$:
But the sixth case repeats the fifth under the replacement $z=x+y,$ and the eighth repeats the seventh under the replacement $z=x+y+1.$
The first ring is $\mathbb{F}_2[z]/(z^3),$ where the generator of the $\mathbb{F}_4$ subring is $z+1$ and $y=z^2.$ The second is clearly $\mathbb{F}_4\times\mathbb{F}_2.$ The third is $\mathbb{F}_2[x,y]/(x^2,xy,y^2)$ verbatim, the fifth is $\mathbb{F}_2[x]/(x^2)\times\mathbb{F}_2,$ and the fourth is $\mathbb{F}_2[y]/(y^3).$ The seventh repeats the fifth under $x\leftrightarrow y,$ the ninth is clearly $\mathbb{F}_2^3,$ and the last is $\mathbb{F}_4\times\mathbb{F}_2,$ where $(\omega,0)=y$ and $(0,1)=x.$
If $xy$ is one of the "upper four," then all possibilities for the value of $xy$ in the cases of the subrings $\mathbb{F}_4$ or $\mathbb{F}_2[1+i]$ lead to contradictions with the associativity of $x^2y$ (here $x=\omega,1+i$ resp.)
In the case of $\mathbb{F}_2^2,$ if $xy$ is one of the "upper four" and $y^2$ is one of the "lower four," then we can say WLOG either $y^2=0$ or $y^2=x.$ This makes a priori 8 possibilities. But only the following two do not lead to contradictions:
(I have written $\mathbb{F}_2^2$ using an idempotent $e=x.$)
The former is $\mathbb{F}_2[i]\times\mathbb{F}_2,$ where $y=(1+i,0)$ and $x=(0,1),$ and the latter is the former under the transformation $z=x+y.$
If $xy$ and $y^2$ are in the "upper four," we still have a priori 16 cases to consider. But in fact we must have either $xy=y$ or $xy=x+y+1$ for the associativity of $x^2y,$ thus there are only eight.
($x$ is an idempotent nonunit of $\mathbb{F}_2^2$ in each case.)
But the second and fourth violate the associativity of $xy^2.$ I'll leave it to you to determine which of the first, third, and fifth through eighth are valid rings, and which rings in your list they correspond to.
Second Proof
This begins as a converse to what you noted. We have either a field, a ring with a unique nonzero maximal ideal, or a nonlocal ring. In the case of a field we know it's $\mathbb{F}_8.$ In the case of a nonlocal ring we have Theorem 3.1.4 on page 40 of Finite Commutative Rings and Their Applications by Bini and Flamini:
(It's a good read!) and so this leaves only the local case.
The maximal ideal $\mathfrak{m}$ has either two or four elements, and we'll rule out $|\mathfrak{m}|=2$ at the end.
In the case of four elements, let's call those elements $0,x,y,x+y.$ If $\mathfrak{m}^2=0$, we have $x^2=xy=y^2=0,$ so we get $\mathbb{F}_2[x,y]/(x^2,xy,y^2).$
If $\mathfrak{m}^2\neq0,$ then we can't have both $x^2=0$ and $y^2=0,$ or else $(x+y)^2=0,$ and thus $(xy)^2=0$ leads to a contradiction in any of the three choices $xy=x,~xy=y,~xy=x+y$ by multiplying through by $x$ or $y.$ But $xy\in\mathfrak{m}$ then implies $xy=0,$ which means $\mathfrak{m}^2=0,$ contradiction. So WLOG $y=x^2,$ our eight elements are the at-most-quadratic polynomials in $x,$ and we need only determine which of the four elements of $\mathfrak{m}$ is $x^3.$ The choice $x^3=0$ leads to $\mathbb{F}_2[x]/(x^3).$ The choices $x^3=x$ and $x^3=x^2$ lead to $x+1$ being uninvertible, which is a contradiction since any element outside of $\mathfrak{m}$ must be invertible. And the choice $x^3=x^2+x$ leads to $x^2+x+1$ being uninvertible.
Finally, if $|\mathfrak{m}|=2,$ then $R/\mathfrak{m}\cong\mathbb{F}_4$ and $R$ has six units. Take a unit $x$ other than 1. Then $\bar x=\omega$ by an isomorphism, $\bar x^2=\omega+1,$ and $\bar x^3=1,$ and $x,x^2,x^3$ are all units in $R.$ Say $\mathfrak{m}=\{0,y\}.$ We can't have $y=x+1$ because they go to different things in the quotient, and so we have a unique way to write all eight elements of $R.$ Now $xy\in\mathfrak{m}$ implies $xy=0$ or $xy=y.$ But note too that $(x+1)y=0$ or $(x+1)y=y.$ Only two of the four possibilities do not lead directly to a contradiction and they are symmetric, so we may say $xy=0.$ But then if $x^2=x+1,$ multiplying through by $y$ gives us $y=0,$ and if $x^2=x+y+1$ then multiplying through by $y$ gives us $y^2=y.$ These are the only two possibilities since $\bar x^2=\overline{x+1},$ the former is a contradiction since $y\neq0,$ and the latter gives us $R\cong\mathbb{F}_4\times\mathbb{F}_2,$ which is a contradiction since we said $R$ is local.