I have a proposition, maybe you can find some weak points in it:
It's based on Elo. First, all the players get 1000 rating points ($R$). Let's focus on 1 vs 1, in the end I will show how to deal with 2 vs 1 and 2 vs 2. Player's A expected score $E_A$ is calculated as follows:
$$ E_A = \frac{1}{1+10^{(R_B - R_A)/500}} \in (0, 1)$$
Where $R_A$ is a point rating for player A. Player's B expected rating is $E_B = 1 - E_A$.
New rating for player A:
$$R_A' = R_A + 100(S_A - E_A)$$
Where $S_A$ is the actual score. In case of winning $10:i$ $(i \in {0, 1, ..., 9}$) it is1:
$$S_A = \frac{10}{10+i}$$
In case of losing $i:10$ it is:
$$S_A = 1-\frac{10}{10+i}$$
Example:
Player A with $R_A = 1000$ points won 10:6 with player B with $R_B = 1000$ points. In such case, expected score $E_A = E_B = 0.5$. Actual score $S_A$ for player A is $\frac{10}{10 + 6}=0.625$. So, $R_A' = 1000 + 100(0.625 - 0.5) = 1000 + 12.5 = 1012.5$, $R_B' = 1000 - 12.5 = 987.5$.
What to do with 2 players A and B in one team playing versus player C? We can treat player A and B as one player with rating $\frac{R_A + R_B}{2}$. So, $E_A = E_B = \frac{1}{1+10^{(R_C - \frac{R_A + R_B}{2})/500}}$. New rating will be calculated individually:
$$R_A' = R_A + 100(S_{AB} - E_A), R_B' = R_B + 100(S_{AB} - E_B)$$
$S_{AB}$ is the actual game score for player A and B. Since $E_A = E_B$, rating change for players in the same team is the same.
2 vs 2 case analogically.
1Why $S_A = \frac{10}{10+i}$?
Consider a winning $10 : i$. Let $p$ be the probability of the winning team scoring a goal. This will be our "actual score (performance)". Let's assume that this probability is constant for the whole game. After seeing the actual outcome, $10 : i$, we can ask: what's the most likely $p$? One can use maximum likelihood estimation. What's the probability that it was $10:i$, given that the probability of the winning team scoring a goal was $p$? It is:
$P(10:i | p) = \binom{9+i}{9}p^{10} (1-p)^i$
There were $10 + i$ goals in total. The last one belong to the winning team, it was their 10th goal, so this one is pinned. $\binom{9+i}{9}$ is the number of 9 winners goals and $i$ losers goals combinations. Multiplied by the probability of winners scoring 10 goals, $p^{10}$.
Multiplied by the probability of losers scoring $i$ goals, $(1-p)^i$. Their goals positions are already determined, so no combinations here.
Now we just need to find which $p$ maximizes $P(10:i | p)$. One could just take the first derivative and set it to zero. However, the maximum is the same for $\log P(10:i | p)$ and it will be easier to find maximum for it, for log-likelihood:
$\log P(10:i | p) = \log \binom{9+i}{9} + \log p^{10} + \log (1-p)^i = \log \binom{9+i}{9} + 10 \log p + i \log (1-p)$
Set first derivative of $P$ with respect to $p$ to 0:
$\log P(10:i | p)' = \frac{10}{p} + \frac{i}{p-1} = 0$
$\frac{10(p-1) + ip}{p(p-1)} = 0$
$10p-10 + ip = 0$
$p(10+i) = 10$
$p = \frac{10}{10+i}$
I believe, if you want to classify the players so each one will have matches with someone with a similar "competence", the average won't help if they aren't playing all the matches, but neither the total amount of winnings, and neither the amount of winnings over effectively played matches, since, because of the reduced number of samples, every "outlier" will "weight" "too much" in these "metrics".
Just as an idea, maybe ordering them using their probability of winning the next match, using Laplace's Rule of Succession [1], could somehow "balance" how much winnings they have with how many matches they have played:
$$ P(\text{player}_i\,\text{winning next match}) = \frac{\text{n° matches winned by}\,\text{player}_i + 1}{\text{n° matches played by}\,\text{player}_i + 2}$$
you can multiply it later by a hundred to make this "below one" numbers "look better" as earn points - it actually create a moving relative scale.
And also, each player probabilities could be being updated through the same rule or maybe using Bayes' Rule [2], or intruding "odds" and opening the window to "bets" hahaha.
Hope this will help you, I your position, I would be creating an Excel spreadsheet and comparing the different alternatives on a table, to compare which one fulfill better the club requirements.
PS: for new players, using this probability ordering, you will have at the beginning a probability of 1/2 instead of starting from 0 (as if they know nothing about tennis). This because you have no information about their skills. This is also a good thing, since you will never have someone like Roger Federer playing with a total mess like me (making me sad about this league, or getting Roger bored), and also, people that is starting in the "middle class" will fastly converge into their proper category.
Best Answer
I've recently being working on this for a system to rate our (highly competitive) office table football. I found a solution in a slightly strange post featuring an "interview" of the Elo algorithm.
Essentially, you apply a multiplier to the K factor to account for margin of victory which is calculated as:
Where PD is the point differential in the game, ELOW is the winning team’s Elo Rating before the game, and ELOL is the losing team’s Elo Rating before the game.
It works quite nicely due to the inclusion of the natural algorithm, since large wins are given more weighting but very large wins don't affect the ratings too greatly.