Extensions: Spectrum Problem
Given a C*-algebra $\mathcal{A}_0$ and unital extensions $1\in\mathcal{A}$ and $1'\in\mathcal{A}'$.
Regard a common element:
$$A_0\in\mathcal{A}_0:\quad A^{(\prime)}:=\iota^{(\prime)}(A_0)$$
Can it happen that its spectra really differ:
$$\sigma(A)\cup\{0\}\neq\sigma(A')\cup\{0\}$$
(The scenario is inspired by possible noncanonical unital extensions.)
One might be tempted to conclude that they agree as:
$$\langle\iota(\mathcal{A}_0)\cup\{1\}\rangle\cong\mathcal{A}_0\oplus\mathbb{C}\cong\langle\iota'(\mathcal{A}_0)\cup\{1'\}\rangle'$$
But there is a flaw as the example below illustrates.
Example
Given the matrix algebra $\mathcal{M}:=M(2;\mathbb{C})$.
Consider the strict subalgebra:
$$\mathcal{M}':=\left\{\begin{pmatrix}a&0\\0&0\end{pmatrix}:a\in\mathbb{C}\right\}\subseteq\mathcal{M}$$
Then their units disagree:
$$1=\begin{pmatrix}1&0\\0&1\end{pmatrix}\neq\begin{pmatrix}1&0\\0&0\end{pmatrix}=1'$$
Regard their common element:
$$M:=\begin{pmatrix}1&0\\0&0\end{pmatrix}\in\mathcal{M}\cap\mathcal{M}'$$
Then it has at least distinct spectra:
$$\sigma(M)=\{0,1\}\neq\{1\}=\sigma(M)'$$
(Still, leaving open the question wether they really can differ.)
 A: Disclaimer
There was another answer giving the motivation for this one. Sadly, it got deleted.
Crucial Point

It boils down to the existence of an isomorphism: $\pi[A]=A'$ & $\pi[1]=1'$
This is ill-defined only for the case: $1\propto A$ & $1'!\propto A'$

Prework
Suppose proportionality holds: $1=\lambda A$
Then the original algebra was unital: $1_0=\lambda A_0\in\mathcal{A}_0$
Pathological Case
Assume proportionality is trivial too: $\lambda=0$
So one checks: $(A=0=1)$ & $\left(A'=0'\neq1'\right)$
Thus the spectra differ by zero:
$$\sigma(A)=\varnothing\approx\{0\}=\sigma(A')$$
(Besides, the original algebra was pathological then!)
Proper Case
Assume proportionality is nontrivial only: $\lambda\neq0$
So one checks: $\left(A=\frac{1}{\lambda}1\right)$ & $(1\neq0)$ & $\left(A'=\frac{1}{\lambda}E'\right)$ & $\left(0'\neq E'\neq1'\right)$ & $\left(E^2=E=E^*\right)$
Thus the spectra differ by zero:
$$\sigma(A)=\{\frac{1}{\lambda}\}\approx\{0,\frac{1}{\lambda}\}=\sigma(A')$$
(Besides, the original algebra was unital then!)
Standard Case
In any other case there exists an isomorphism: $\pi[1]=1'$
Thus the spectra agree by spectral permanence:
$$\sigma(A)=\sigma(\pi[A])=\sigma(A')$$
(Besides, it didn't matter wether the algebra was pathologic or unital here!)
Key Observation

Observe that the spectra only can differ if the original algebra was unital: $A_0\propto1_0\in\mathcal{A}_0$

Thus one can unambigously define the spectrum by any unital extension!
