No.
Throwing two dice is the same as throwing a dice twice.
Let's think naturally:
What is the set of outcomes when a die is thrown twice?
It is just the set of all ordered pairs such that, the first of the two entries tell you the outcome on the first throw and the second of the entries tell you the outcome on the second throw.
What is the set of outcomes when two dice are thrown?
Without loss of generality, I can label the dice as die A and die B. So, now, the set of outcomes will be the set of ordered pairs such that he first of the two entries tell you the outcome on die A and the second of the entries tell you the outcome on the die B.
So, radically, the set of outcomes, technically called the sample space, is the same.
To conduct an experiment with $100$ dice, if you have the means, you could buy $100$ dice and use a method that is reasonably unbiased in generating outcomes, save time or throw a single die $100$ times, the outcomes will be unbiased but you'll waste time.
The point is both methods have their own practical merits and demerits.
The $\frac16$ and $\frac13$ should be squared at first, not doubled at the end. The correct calculation for $P(S)$ is
$$P(S)=\frac45\cdot\frac1{6^2}+\frac15\cdot\frac1{3^2}=\frac1{45}+\frac1{45}=\frac2{45}$$
The second term above is $P(L\cap S)$, so
$$P(L\mid S)=\frac{P(L\cap S)}{P(S)}=\frac{1/45}{2/45}=\frac12$$
Best Answer
Since $1+2+\cdots+6=21$, the probabilities of $1,2,3,4,5,6$ in $1$ toss are respectively $\frac{1}{21}$, $\frac{2}{21}$, and so on up to $\frac{6}{21}$.
The probability of a sum of $2$ in $2$ throws is $\frac{1}{21}\cdot\frac{1}{21}$.
The probability of a sum of $3$ is $2\cdot \frac{1}{21}\cdot\frac{2}{21}$.
The probability of a sum of $4$ is $2\cdot \frac{1}{21}\cdot\frac{3}{21}+\frac{2}{21}\cdot\frac{2}{21}$.
Add up.
Remark: Your numerator, and therefore the basic analysis, was right. The denominators were not.