Note: I have given some details on the Terry-Bradley model in another answer, which I think is a more effective approach. This is a more direct answer, which is slightly too long for a comment.
The problem with ELO is, even though it is easy to compute ELO scores for your problem, there is no easy way to get a "standard deviation" for each player, like you request. The rating system, however, could go something like this:
A win is worth a score of $1$ and a loss is worth $0$. Let $R_i$ be the current rating of player $i$ (suppose each player starts at $600$, or something). When team $(i,j)$ plays team $(k,\ell)$, we can compute the expected score for the first team as
$$E_{i,j} = \frac{1}{1+10^{(R_k+R_\ell - R_i-R_j)/400}}$$
and for the second team as $E_{k,\ell} = 1 - E_{i,j}$.
Let's suppose that team $(i,j)$ beats $(k,\ell)$ so the observed scores are $S_{i,j} = 1$ and $S_{k,\ell}=0$. The difference between the observed and expected scores is given by
$$D_{i,j} = 1 - E_{i,j}, \qquad D_{k,\ell} = - E_{k,\ell}.$$
Now we need to raise the rating of players $i$ and $j$ and decrease the rating of players $k$ and $\ell$. Let's use the formulas:
$$\begin{aligned}
R^\text{new}_i &= R^\text{old}_i + K_i(1 - E_{i,j}) & \text{rating goes up} \\
R^\text{new}_j &= R^\text{old}_j + K_j(1 - E_{i,j}) & \text{rating goes up} \\
R^\text{new}_k &= R^\text{old}_k - K_k E_{k,\ell} & \text{rating goes down} \\
R^\text{new}_\ell &= R^\text{old}_\ell - K_\ell E_{k,\ell} & \text{rating goes down}. \\
\end{aligned}$$
The $K_i$ terms determine how much the rating changes and also how much "relative credit" player $i$ vs. player $j$ should get for the outcome. One option is to set $K_i$ to a constant, such as $16$ or $32$ as in chess. But another option is to say that the stronger player should get more credit for a win, but also take a larger penalty for a loss. In this scenario, we could set
$$K_i = K_0\frac{R_i^\text{old}}{R_i^\text{old} + R_j^\text{old}},$$
where $K_0$ is a constant (like $16$ or $32$) that you will have to play around with. Larger values mean that ratings will change more quickly.
Update: Questions in Comment
Is there a way to include score so that big wins adjust more than close ones?
Yes. The observed scores $S_{ij}$ and $S_{k\ell}$ should be numbers between $0$ and $1$, but there is wiggle room with how you define them. For instance, you could set
$$S_{ij} = \frac{\# \text{ of sets won by team $(i,j)$}}{\# \text{ of sets won by team $(i,j)$} + \# \text{ of sets won by team $(k,\ell)$}}.$$
So if team $i,j$ wins 2 out of 3 sets they get $S_{ij}=2/3$ and the losing team gets $S_{kj}=1/3$. This approach may need to be adjusted if you want to use points instead of sets, because the winning team score will be too close to $0.5$.
A word of caution: With this approach, it is possible that the winning team can actually lose rating points. This is like how a stronger player loses points for a draw in chess. If you want to avoid this, you could also incorporate this into the $K$ terms by using a larger value of $K$ for a blowout and a smaller value for a close game (e.g. $K = 16 + 2\times \text{|Point Differential|}$).
Is there a way to do an approximation or estimate for updating the sigma/confidence?
Not easily. The notion of confidence doesn't make a lot of sense in traditional ELO. For this, you need a probability model. See my other answer for details.
