Consider a $9\times9$ array, where the row represents the event and the column represents the round. For each house, arrange the $9$ copies numbers from $1$ to $9$ in the array, so that each number appears once in each row and each column. That is, make a $9\times9$ Latin square. Call these arrays $(a_{ij}),(b_{ij}),(c_{ij}),(d_{ij})$ for groups A,B,C,D, respectively.
If $a_ij=m$ for example, it means that team $A_m$ will play event $i$ in round $j$. Now suppose that arrays $(a_{ij}),(b_{ij})$ are orthogonal, that is if $$a_{ij}=a_{mn}=x$$ and $$b_{ij}=b_{mn}=y$$ then $$x=y.$$ Then $A_x$ meets team $B_y$ at most once, and since $A_x$ must meets $9$ different teams, it meets team $B_y$ exactly once.
We see that the conditions can be fulfilled if and only if there exist $4$ pairwise orthogonal Latin squares. Since $9=3^2$ is a prime power, it is known that there exist $8$ pairwise orthogonal Latin squares, and the answer is "yes."
The construction is described here.
EDIT
On reflection, I think I should have included a specific construction. Here is a set of $8$ mutually orthogonal $9\times9$ Latin squares, the maximum possible.
$$\begin{array}{|c|c c c|c c c|c c c|}\hline&1&2&3&4&5&6&7&8&9\\ \hline 1 &11111111&56489723&98765432&69358247&72634859&24973568&85296374&37542986&43827695\\
2 &22222222&64597831&79846513&47169358&83415967&35781649&96374185&18653794&51938476\\
3 &33333333&45678912&87954621&58247169&91526748&16892457&74185296&29461875&62719584\\
\hline4 &44444444&89723156&32198765&93682571&15967283&57316892&28539617&61875329&76251938\\
5 &55555555&97831264&13279846&71493682&26748391&68124973&39617428&42986137&84362719\\
6 &66666666&78912345&21387954&82571493&34859172&49235781&17428539&53794218&95143827\\
\hline7 &77777777&23156489&65432198&36925814&48391526&81649235&52863941&94218653&19584362\\
8 &88888888&31264597&46513279&14736925&59172634&92457316&63941752&75329461&27695143\\
9 &99999999&12345678&54621387&25814736&67283415&73568124&41752863&86137542&38476251\\
\hline\end{array}$$
The rows represent rounds, and the columns represent events. The table is for $8$ houses, say $A,B,C,D,E,F,G,H.$ The sequence indicates which teams from each house will compete in a particular event in a given round. For example, at row $3$, column $4$ we find $58247169$. The interpretation of this is that in round $3$, the teams that compete in event $4$ are $$A_5,B_8,C_2,D_4,E_7,F_1,G_6,\text{ and }H_9$$
In the instant case, where there are only $4$ houses, the last $4$ digits or each table entry can simply be ignored.
I don't think this can work if you have an even number of games. (Wrong: see edit below.) But if you have 5 or 7 games (with 10 or 14 teams), then Chas Brown's solution works perfectly. Here is the seating plan for 5 games (the game stations are named A through E, and the teams are numbered 1 to 10):
A B C D E
-------------
Round 1: 1 2 3 4 5
6 7 8 9 10
Round 2: 2 3 4 5 1
10 6 7 8 9
Round 3: 3 4 5 1 2
9 10 6 7 8
Round 4: 4 5 1 2 3
8 9 10 6 7
Round 5: 5 1 2 3 4
7 8 9 10 6
This is the problem of designing pairings for duplicate bridge tournaments, where instead of different games, we have different deals. See this link for a discussion of the different strategies; and note that the Mitchell Movement with an even number of tables ends up with each pair having to skip one of the deals.
Edited to add: Especially Lime comments that a "perfect schedule" is possible when the number of games is a multiple of four (the so-called double-weave Mitchell Movement). Here is an example with eight games:
A B C D E F G H A B C D E F G H
---------------------- ----------------------
Round 1 1 2 3 4 5 6 7 8 Round 2 2 1 4 3 6 5 8 7
9 10 11 12 13 14 15 16 16 11 10 13 12 15 14 9
Round 3 7 4 1 6 3 8 5 2 Round 4 4 7 6 1 8 3 2 5
11 16 13 10 15 12 9 14 14 13 16 15 10 9 12 11
Round 5 5 6 7 8 1 2 3 4 Round 6 6 5 8 7 2 1 4 3
10 9 12 11 14 13 16 15 15 12 9 14 11 16 13 10
Round 7 3 8 5 2 7 4 1 6 Round 8 8 3 2 5 4 7 6 1
12 15 14 9 16 11 10 13 13 14 15 16 9 10 11 12
Edited again to add: As Chas Brown points out in the comments, this problem is just another disguise for Graeco-Latin squares, which means we can construct perfect schedules for all sizes except $2$ and $6$. This paper by D.A. Preece and B.J. Vowden has examples of $10\times 10$ Graeco-Latin squares.
Best Answer
Using the constraint solver MiniZinc, I arrived at the following:
As a bonus beyond what was asked for in the question, I added constraints to prevent that teams have to play the same event in the next round. Also, additional constraints make sure that each team plays against all other teams, but the teams do not play against the same team in the next round.
The solution is found in under two seconds.
The MiniZinc code: