1. Identify the given elements and their properties:
- Two identically oriented equilateral triangles $\triangle ABC$ and $\triangle A'B'C'$ with centers $S$ and $S'$ respectively.
- $A' \neq S$ and $B' \neq S$.
- $M$ is the midpoint of $A'B$.
- $N$ is the midpoint of $AB'$.

2. **Prove that $\triangle SMB'$ is a 30-60-90 triangle:**
- Extend $BS$ to $D$ such that $S$ is the midpoint of $BD$.
- Extend $A'B'$ to $E$ such that $B'$ is the midpoint of $A'E$.

3. **Show that $\triangle BCD$ and $\triangle ECA'$ are 30-60-90 triangles:**
- Since $\triangle ABC$ and $\triangle A'B'C'$ are equilateral, the angles in these triangles are all $60^\circ$.
- By construction, $S$ is the midpoint of $BD$, making $\triangle BCD$ a 30-60-90 triangle.
- Similarly, $B'$ is the midpoint of $A'E$, making $\triangle ECA'$ a 30-60-90 triangle.

4. Use spiral symmetry to show similarity:
- By spiral symmetry, $\triangle BCE \sim \triangle DCA'$.
- Since $EC \perp A'C$, we have $DA' \perp BE$.

5. Calculate the ratio of the sides:
- Since $\triangle BCD$ and $\triangle ECA'$ are 30-60-90 triangles, the ratio of their sides is $\frac{1}{\sqrt{3}}$.
- Therefore, $\frac{DA'}{BE} = \frac{A'C}{CE} = \frac{1}{\sqrt{3}}$.

6. **Prove that $SM \parallel DA'$ and $MB' \parallel BE$:**
- Since $SM \parallel DA' \perp BE \parallel MB'$, we have $\frac{SM}{MB'} = \frac{DA'}{BE} = \frac{1}{\sqrt{3}}$.

7. **Conclude that $\triangle SMB'$ is a 30-60-90 triangle:**
- Since $\frac{SM}{MB'} = \frac{1}{\sqrt{3}}$, $\triangle SMB'$ is a 30-60-90 triangle.

8. **Similarly, prove that $\triangle SNA'$ is a 30-60-90 triangle:**
- By similar arguments, $\triangle SNA'$ is also a 30-60-90 triangle.

9. **Conclude that $\triangle SB'M$ and $\triangle SA'N$ are similar:**
- Since both $\triangle SMB'$ and $\triangle SNA'$ are 30-60-90 triangles, they are similar by definition.

$\blacksquare$