Transformation of normal distribution Suppose I have a normal distribution $A \sim \mathcal{N}(\mu,\sigma^{2})$ with a known cuttof point (percentile) on this distribution called $o$. Based on this point the distribution needs to be transformed which leads to a new distribution called $W$. Next, suppose $s$ equals a deterministic value. $o$ and $s$ are known in advance and do not need to be estimated.
 The first part of $A \in [-\infty,o]$ becomes deterministic and equal to $o+s$, the second part $]o,+\infty]$ remains stochastic and is equal to a transformation of $A$ more specifically $\mathcal{N}(\mu+s,\sigma^{2})$
So summarised:
$W = \begin{cases}o+s & \text{if } A \leq o \\ A+s & \text{if } A > o \end{cases}$
I want to find an expression for the pdf of $W$.
More specifically in my application A represents the arrival time distribution at a customer and $o$ represent the opening time at that customer. If you arrive to early you will have to wait before you can serve (service time is equal to $s$) the customer.
My current (unsuccesfull) solution approach for the case in which $\mu=10,\sigma=1,o=11,s=1$ is as follows
I think W is the component mix:
$W= P(A\leq o)*(o+s)+ P(A>o)*(A+s),$
I entered this in mathematica (see picture below)
Furthermore, I programmed a monte carlo simulation executing 100000 iterations. The new distribution has a mean equal to $12.08$ and a standard deviation equal to $0.26$.


 A: I can't bring myself to use an 'a' or $A$ for a random variable (or an 'o' for a constant), so will swap for $X$ and $c$.
The problem
Let $X \sim N(\mu,\sigma^2)$ with pdf $f(x)$, and let $c$ and $s$ denote constants.

(source: tri.org.au)
Then, random variable $W$ has a censored distribution, where:
$$W = \begin{cases}c+s & \text{if } X \leq c \\ X+s & \text{if } X > c \end{cases}$$

(source: tri.org.au)
The tail of $X$ to the left of $c$ becomes the discrete mass at $c+s$, with mass $P(X \leq c)$ (the blue point). The tail of $X$ to the right of $c$ gets shifted to $X+s$.

As to computation of $E[W]$:
Define:

(source: tri.org.au)
Then, $E[W]$ is:

(source: tri.org.au)
where I am using the Expect function from the mathStatica package for Mathematica to automate the computation, and Erf denotes the error function.
For the given parameter values:
{$\mu$ -> 10, $\sigma$ -> 1, c -> 11, s -> 1}
... this yields:  $\frac{1}{2} \left(\text{erf}\left(\frac{1}{\sqrt{2}}\right)+23\right)+\frac{1}{\sqrt{2 e \pi }} \approx 12.0833$ as per OP's simulation.
Suggestion
The inclusion of constant $s$ is an unnecessary distraction. A neater way of modelling this is to define:
$$Y = \begin{cases}c & \text{if } X \leq c \\ X & \text{if } X > c \end{cases}$$
and $$W = Y + s$$
