What is the effect of noise on Shannon entropy?
The grayscale image is actually a discrete image. entropy calculates a histogram and from that extracts "empiric" probabilities, to be used in the common information formula

$$I= -\log_2 (P(g))$$

with $P(g)$ being the (empiric) probability of an intensity value $g$.

Now, entropy $H$ is the expectation of Information. For your whole image $V$ that means 

$$\begin{align}
H(V) &amp;= E\left\{ I(g_{x,y})\right\}\\
&amp;= \sum\limits_{x,y\in \text{idx(V)}} P(g_{x,y})\cdot\left(-\log_2 (P(g_{x,y}))\right)\\
&amp;= -\sum\limits_{x,y\in \text{idx(V)}} P(g_{x,y})\log_2 (P(g_{x,y}))
\end{align}$$

If you look at the product in the sum intensely enough, you'll notice that the sum will take a maximum value if all properties are equal – that is, if every intensity value is just as likely as every other. The histogram of that image would be a flat line.

Adding "well-behaved" noise will make almost any image look more like that – in the end, pixels with identical intensity will be relatively common in most natural images, and adding noise will "break" that.

Now, salt'n'pepper is different: you add extreme values to your image. That will make the intensity at many pixels clip at the maximum intensity (or the minimum one), skewing the histogram away from flat to a very "horned" thing, which has less entropy.