You’re certain to get at least one correct guess, and in many cases you’ll get more than one, so the expected number of correct guesses is clearly more than one. You’ll get a second correct guess whenever you draw the $1$ before you draw the $2$; the probability of that is $\frac12$, so the expected number of correct guesses can already be seen to be at least $\frac12\cdot1+\frac12\cdot2=\frac32$.
In general you’ll get at least $n$ correct guesses if the numbers $1,\dots,n$ appear in their correct order, ignoring any larger numbers that may appear between them. The probability of that is $\frac1{n!}$, since all $n!$ orders in which they could appear are equally likely. The probability that $n+1$ is the last of the set $\{1,\dots,n+1\}$ to be drawn is $\frac1{n+1}$, so the probability that it is drawn before you draw all of the smaller numbers is $\frac{n}{n+1}$.
Thus, the probability that you’ll get exactly $n$ correct guesses is $$\frac1{n!}\cdot\frac{n}{n+1}=\frac{n}{(n+1)!}\;,$$ and the expected number of correct guesses is
$$\sum_{n=1}^{100}\frac{n}{(n+1)!}\cdot n=\sum_{n=1}^{100}\frac{n^2}{(n+1)!}\;.$$
Now $$e^x=\sum_{n\ge 0}\frac{x^n}{n!}\;,$$ so $$\frac{e^x-1}x=\sum_{n\ge 1}\frac{x^{n-1}}{n!}=\sum_{n\ge 0}\frac{x^n}{(n+1)!}\;,$$
$$x\frac{d}{dx}\left(\frac{e^x-1}x\right)=\sum_{n\ge 0}\frac{nx^n}{(n+1)!}\;,$$ and
$$\frac{d}{dx}\left(x\frac{d}{dx}\left(\frac{e^x-1}x\right)\right)=\sum_{n\ge 0}\frac{n^2x^{n-1}}{(n+1)!}\;.\tag{1}$$
Finally, $$\frac{d}{dx}\left(x\frac{d}{dx}\left(\frac{e^x-1}x\right)\right)=\frac{d}{dx}\left(\frac{xe^x-e^x+1}x\right)=e^x-\frac{(x-1)e^x}{x^2}-\frac1{x^2}\;.\tag{2}$$
Evaluating the righthand sides of $(1)$ and $(2)$ at $x=1$, we see that
$$\sum_{n\ge 1}\frac{n^2}{(n+1)!}=e-1\;,$$ so the expected number of correct guesses is just a hair less than $e-1\approx 1.718281828459045$.
