You forgot that you can have strings with two copies of one letter, two of another, and one of the third. There are $3$ ways to choose which of the letters is used only once, and $5$ ways to place it in the string. There are then $\binom42=6$ ways to choose which two of the remaining four places get the two copies of the first of the other two letters in alphabetical order, for a total of $3\cdot 5\cdot 6=90$ strings, exactly the number that you’re missing.
Added: To do it with generating functions, let $a_n$ be the number of words of length $n$ over a $3$-letter alphabet that use each letter at least once. The first step is to find a recurrence for $a_n$. For brevity call such words good. Each good word of length $n-1$ can be extended to a good word of length $n$ in $3$ different ways, since each of the three letters can be added at the end of the word. In addition, each $(n-1)$-letter word that contains exactly two of the three letters can be extended to a good word in one way, by adding the missing letter. How many of these $(n-1)$-letter words are there? There are $3$ ways to choose which of the three letters is missing, and each of the $n-1$ positions in the word can contain either of the two remaining letters. However, we have to exclude the two possibilities that contain only one of the two letters, so there are $3(2^{n-1}-2)$ such words. This leads us to the recurrence
$$a_n=3a_{n-1}+3(2^{n-1}-2)\tag{1}\;.$$
Clearly $a_0=a_1=a_2=0$ and $a_3=3!=6$. However, if we assume that $a_n=0$ for $n<0$, $(1)$ requires a small adjustment to get $a_0$ and $a_1$ right: as it stands, it yields the values $-9/2$ and $-3$. To correct this, we add a couple of Iverson brackets to get
$$a_n=3a_{n-1}+3(2^{n-1}-2)+\frac92[n=0]+3[n=1]\tag{2}\;.$$
Now multiply through by $x^n$ and sum over $n$ to get the generating function $f(x)$:
$$\begin{align*}f(x)=\sum_na_nx^n&=3\sum_n\left(a_{n-1}+2^{n-1}-2+\frac32[n=0]+[n=1]\right)x^n\\
&=3\left(\sum_na_{n-1}x^n+\sum_n2^{n-1}x^n-2\sum_nx^n+\frac32+x\right)\\
&=3\left(x\sum_na_{n-1}x^{n-1}+\frac12\sum_n(2x)^n-\frac2{1-x}+\frac32+x\right)\\
&=3\left(x\sum_na_nx_n+\frac{1/2}{1-2x}-\frac2{1-x}+\frac32+x\right)\\
&=3xf(x)+\frac32\left(\frac{7x-3}{(1-x)(1-2x)}+2x+3\right)\\
&=3xf(x)+\frac{6x^3}{(1-x)(1-2x)}\;,
\end{align*}$$
so $$(1-3x)f(x)=\frac{6x^3}{(1-x)(1-2x)}\;,$$ and
$$\begin{align*}
f(x)&=\frac{6x^3}{(1-x)(1-2x)(1-3x)}\\
&=x^3\left(\frac3{1-x}-\frac{24}{1-2x}+\frac{27}{1-3x}\right)\\
&=3x^3\left(\sum_nx^n-8\sum_n2^nx^n+9\sum_n3^nx^n\right)\\
&=\sum_n\left(3-3\cdot 2^{n+3}+3^{n+3}\right)x^{n+3}\\
&=\sum_n\left(3-3\cdot 2^n+3^n\right)x^n\;.
\end{align*}$$
Equating coefficients, we finally have $a_n=3^n-3\cdot 2^n+3$.
You are getting the right answer, but your explanation of it doesn't sound quite right. I would approach it this way:
First, pick two letters not to use. This can be done in ${5\choose2}$ ways. Next, with the remaining three letters, consider the two cases:
Pick one letter to use three times, the other two being used just once. This can be done $3$ ways, and then the letters can be arrange in $5\choose1,1,3$ ways.
Pick one letter to use just once, with the other being being used twice each. Again, there are $3$ choices for the singlet, and then $5\choose1,2,2$ ways to arrange the letters.
The total number of words with three different letters is thus
$${5\choose2}\left(3{5\choose1,1,3}+3{5\choose1,2,2}\right)=10(3\cdot20+3\cdot30)=1500$$
Best Answer
Your mistake is in the arithmetic. What you think comes out to 2341 really does come out to 240.
$4^5=1024$, $3^5=243$, $2^5=32$, $1024-(4)(243)+(6)(32)-4=1024-972+192-4=1216-976=240$