If $Y$ is lognormally distributed, what is $E|Y - 1|$? Let $Y$ be lognormal with parameters $\mu$ and $\sigma$, such that $\log(Y)$ is Gaussian with mean $\mu$ and variance $\sigma^2$.
I know $0 < Y < \infty$, $E(Y) = \exp(\mu+\frac{1}{2}\sigma^2)$, and $Var(Y) = \exp(2\mu +\sigma^{2})[\exp(\sigma^{2})-1]$
Let $R = Y - 1$.
$-1 < R < \infty$, $E(R) = \exp(\mu+\frac{1}{2}\sigma^2) - 1$, and $Var(R) = Var(Y) = \exp(2\mu +\sigma^{2})[\exp(\sigma^{2})-1]$
What is $E(|R|)$?
This is the mean of the "folded" distribution of $R$.
Context:
When you compare a measurement $M$ to a known control $C$, the quantity $\frac{M}{C}$ is often lognormally distributed.  $\frac{M-C}{C} = \frac{M}{C} - 1$ is the "relative difference" or RD.  In the above, $\frac{M}{C} = Y$ and $Y - 1 = R$ is the RD.  $E(R)$ is the "mean relative difference" or MRD.
$|R|$ is the "absolute relative difference" (not my choice of terms) or ARD, and $E(|R|)$ is the "mean absolute relative difference" or MARD.  This quantity is used to summarize the accuracy of a measurement system, especially the accuracy of blood glucose monitors.

Continuing on from the previous answer.
We can derive the distribution function of your random variable, which will then allow us to derive its expectation.
Let's say we have
$$X\sim LN(\mu,\sigma^2)$$
Because we know the distribution of $X$, we know its distribution function, $F_{X}(x)$ where $0\leq x <\infty$.
Now we have that $Y=X-1$. We can define the cumulative distribution function of $Y$ as
$$F_{Y}(y)=\text{Pr}(Y\leq y)=\text{Pr}(X-1\leq y)=\text{Pr}(X\leq y+1)=F_{X}(y+1)$$
where $-1\leq y <\infty$.
Furthermore, we know that $Z=|Y|$. Applying a similar approach we get two cases.

For $-1\leq y<0\quad\Rightarrow\quad Z=-Y\quad\Rightarrow\quad 0\leq z<1$:

$$\begin{align}
F_{Z}(z)&=\text{Pr}(Z\leq z)\\
&=\text{Pr}(-Y\leq z)\\
&=1-\text{Pr}(Y\leq -z)\\
&=1-F_{Y}(-z)\\
&=1-F_{X}(-z+1)
\end{align}$$
$$\begin{align}
f_{Z}(z)&=\frac{d}{dz}F_{Z}(z)=f_{X}(-z+1)
\end{align}$$

For $0\leq y<\infty\quad\Rightarrow\quad Z=Y\quad\Rightarrow\quad 0\leq z<\infty$:

$$\begin{align}
F_{Z}(z)&=\text{Pr}(Z\leq z)\\
&=\text{Pr}(Y\leq z)\\
&=F_{Y}(z)\\
&=F_{X}(z+1)
\end{align}$$
$$\begin{align}
f_{Z}(z)&=\frac{d}{dz}F_{Z}(z)=f_{X}(z+1)
\end{align}$$
We then need to calculate the expectation of this random variable:
$$\begin{align}
\mathbb{E}[Z]&=\int_{0}^{1}z\cdot f_{X}(-z+1)dz+\int_{0}^{\infty}z\cdot f_{X}(z+1)dz\\
&=\int_{0}^{1}(1-j)f_{X}(j)dj+\int_{1}^{\infty}(i-1)f_{X}(i)di\quad\quad(j=-z+1, i=z+1)\\
&=\int_{0}^{1}f_{X}(j)dj-\int_{1}^{\infty}f_{X}(i)di-\int_{0}^{1}jf_{X}(j)dj+\int_{1}^{\infty}if_{X}(i)di\\
&=2\cdot F_{X}(1)-1-\int_{0}^{1}jf_{X}(j)dj+\int_{1}^{\infty}if_{X}(i)di\\
&=e^{\mu+\sigma^2/2}\Bigg(2\cdot\Phi\bigg(\frac{\mu}{\sigma}+\sigma\bigg)-1\Bigg)-\Bigg(2\cdot\Phi\bigg(\frac{\mu}{\sigma}\bigg)-1\Bigg)
\end{align}$$
where $\Phi(\cdot)$ is the cumulative distribution function of the standard Normal.

The two integrals above (in terms of $i,j$) can each be evaluated by using the two-sided truncated integral:
$$\begin{align}
\int_{a}^{b}\frac{1}{\sigma\sqrt{2\pi}}\exp\bigg(-\frac{(\ln{x}-\mu)^2}{2\sigma^2}\bigg)dx\\
\end{align}$$
Let $y=\ln{x}-\mu\quad\Rightarrow\quad dx=e^{y+\mu}dy$
$$\begin{align}
\Rightarrow e^{\mu+\sigma^2/2}\int_{\ln{a}-\mu}^{\ln{b}-\mu}\frac{1}{\sigma\sqrt{2\pi}}\exp\bigg(-\frac{(y-\sigma^2)^2}{2\sigma^2}\bigg)dy\\
\end{align}$$
Let $z=\frac{y-\sigma^2}{\sigma}\quad\Rightarrow\quad dy=\sigma dz$
$$\begin{align}
&\Rightarrow e^{\mu+\sigma^2/2}\int_{(\ln{a}-\mu-\sigma^2)/\sigma}^{(\ln{b}-\mu-\sigma^2)/\sigma}\frac{1}{\sqrt{2\pi}}\exp\bigg(-\frac{z^2}{2}\bigg)dz\\
&=e^{\mu+\sigma^2/2}\Bigg(\Phi\bigg(\frac{\ln{b}-\mu-\sigma^2}{\sigma}\bigg)-\Phi\bigg(\frac{\ln{a}-\mu-\sigma^2}{\sigma}\bigg)\Bigg)
\end{align}$$
