Case-bashing here is not so bad (and perhaps somewhat necessary since some of the examples come very close to working).
For ease, I will use the following notation
$$\begin{array}{ccccc} & & A1 & A2 & A3 \\ & & \times & B1 & B2 \\ \hline & C1 & C2 & C3 & C4 \\D1 & D2 & D3 & D4 & \\ \hline E1 & E2 & E3 & E4 & E5 \end{array}$$
So far, you have discovered that
$C4 = E5 = 5$ and either $A3$ or $B2$ is $5$ with the other being $3,5$ or $7$. For now, Let us consider these as three separate cases.
Case 1
One of $A3, B2$ is $7$ and the other is $5$.
Since $7 \times 5$ is $35$, we see that this multiplication carries over $3$ to the next decimal place. This makes $C3$ equal to the last digit of $A2 \times B2 + 3$. The only possibilities therefore are
(i) $A2 = B2 = 7$ and $C3 = 2$.
(ii) $A2 = 2$, $B2 = 5$ and $C3 = 3$.
(iii) $A2 = 2$, $B2 = 7$ and $C3 = 7$.
In case (i), we carry over a $5$ to the next multiplication, $A1 \times B2$ where, this time, both digits from the result must be prime. Checking through the four cases for $A1 = 2,3,5,7$, we quickly see that in none of the results do we achieve that so this is a dead end.
In case (ii), we carry over a $1$ to the next multiplication. Again, checking through the four cases, $A1 = 2,3,5,7$ we see that $A1 \times B2 + 1$ cannot have both its digits prime and so this is another dead end.
In case (iii), we also carry over a $1$ to the next multiplication. This time there is exactly one instance where the product gives us a number, both of whose digits is prime, $A1 = 3$ and thus $C1 = C2 = 2$.
Following on from here, we can check through the four cases of $B1$ multiplied by $325$ ($A1-A2-A3$) to check whether the digits $D1, D2, D3, D4$ are prime. We already know it will work for $B1=7$ and $B1=2$ is an obvious dead end as $D4$ would be $0$.
$3\times 325$ will only have three digits and $5 \times 325$ will begin with a $1$ so they are also out. Hence $B1 = 7$ is the only possibility. Thus, so far we have $$\begin{array}{ccccc} & & 3 & 2 & 5 \\ & & \times & 7 & 7 \\ \hline & 2 & 2 & 7 & 5 \\2 & 2 & 7 & 5 & \\ \hline E1 & E2 & E3 & E4 & E5 \end{array}$$ and evaluating the final sum gives $25025$ so, in this case, only $E3=0$ is non-prime.
Case 2
$A3 = B2 = 5$.
Since $5 \times 5$ is $25$, we see that this multiplication carries over $2$ to the next decimal place. This makes $C3$ equal to the last digit of $A2 \times 5 + 2$. We see that, in this case $A2$ can be any of the four possibilities and $C3$ will be prime. However, if we carry over a $1$ to the next product ($A1 \times 5$) then $C2$ cannot be prime, hence there are only two cases which we need to investigate, namely
(i) $A2 = 5$ and $C3 = 7$.
(ii) $A2 = 7$ and $C3 = 7$.
In case (i), we carry over a $2$ to the next multiplication and to get $C1$ and $C2$ both prime, we must have $A1 = 5$ or $7$.
In case (iii), we carry over a $3$ in which case, no choice for $A1$ will make both $C1$ and $C2$ prime.
Therefore, we only have two remaining cases to consider for $A1-A2-A3$ which are $555$ and $755$. In the $555$ case, we can easily see that $B1$ cannot be $2$ and if $B1=3$ then $D1$ would be $1$. If $B1$ is $7$ then $D3=8$ and so this would also not work.
Hence, we must have $B1=5$ and the product looks like this $$\begin{array}{ccccc} & & 5 & 5 & 5 \\ & & \times & 5 & 5 \\ \hline & 2 & 7 & 7 & 5 \\2 & 7 & 7 & 5 & \\ \hline E1 & E2 & E3 & E4 & E5 \end{array}$$ and evaluating the final sum gives $30525$ so, in this case, only $E2=0$ is non-prime.
When $A1-A2-A3$ is $755$ we can see that $B1$ is not $2$. $B1=3$ makes $D3=6$ and $B1=7$ makes $D3=8$ so the only thing that will work here is $B1=5$.
Overall, we have $$\begin{array}{ccccc} & & 7 & 5 & 5 \\ & & \times & 5 & 5 \\ \hline & 3 & 7 & 7 & 5 \\3 & 7 & 7 & 5 & \\ \hline E1 & E2 & E3 & E4 & E5 \end{array}$$ and evaluating the final sum gives $41525$ so, in this case, both $E1$ and $E2$ are non-prime.
Case 3
One of $A3, B2$ is $3$ and the other is $5$.
Since $3 \times 5$ is $15$, we see that this multiplication carries over $1$ to the next decimal place. This makes $C3$ equal to the last digit of $A2 \times B2 + 1$. The only possibilities therefore are
(i) $A2=2$, $B2=3$ and $C3=7$.
(ii) $A2=7$, $B2=3$ and $C3=2$.
In case (i), we carry over nothing to the next multiplication which means we need both digits of the product $A1 \times 3$ to be prime which doesn't happen for any of the possible values of $A1$.
In case (ii), we carry over a $2$ to the next product, which means that both digits of $A3 \times 3 + 2$ must be prime and this only happens for $A1 = 7$, so that $A1-A2-A3$ is $775$.
Now checking through the cases for $B1$ we see that it cannot be $2$ (otherwise $D4=0$), $B1$ cannot be $5$, otherwise, $D2=8$ and $B1$ cannot be $7$, otherwise $D2=4$. These latter cases can also be ruled out as, otherwise, they would have appeared as candidate solutions in one of the first two main cases. Hence, $B1 = 3$ is the only possibility.
Plugging into the equation we have $$\begin{array}{ccccc} & & 7 & 7 & 5 \\ & & \times & 3 & 3 \\ \hline & 2 & 3 & 2 & 5 \\2 & 3 & 2 & 5 & \\ \hline E1 & E2 & E3 & E4 & E5 \end{array}$$ and evaluating the final sum gives $25575$ all of whose digits are prime so this is the only solution.
