Let
$$E := \{x \in \Bbb R^5: x_1, x_2 \in \Bbb R, x_3 = x_1 + x_2, x_4 = 1, x_5 = 2\}$$
be a manifold. Furthermore, let
$$\phi: \Bbb R^2 \rightarrow E$$
be a chart and $\sigma$ the surface-measure. You are allowed to assume that $(E, B(E))$ is a measurable space.
Show that there exists a $c \in \Bbb R$ such that
$$\sigma(A) = c\lambda^2(\phi^{-1}(A))$$
and conclude that $\sigma$ is a measure.
Since the surface measure $\sigma$ is independent from the chart we choose, we define a new chart $\phi: \Bbb R^2 \rightarrow E$ by
$$\phi(x_1, x_2) := \begin{pmatrix} x_1 \\ x_2 \\ x_1 + x_2 \\ 1 \\ 2 \\ \end{pmatrix}.$$
By definition of the surface measure and the integral on manifolds, it follows that (for $X_A$ being the indicator function)
$$\sigma(A) = \int_A 1 d\sigma = \int_E X_A d\sigma = \int_{\Bbb R^2} X_A(\phi(x_1, x_2)) \sqrt {g} \ d\lambda^2(x, y).$$
$g$ refers to the determinant of the gramian matrix of $M$ with respect to $\phi$. A little bit of calculating gives $g = 4$, hence $\sqrt g = 2$, which is a constant of course. Hence, the above integral can be written as
$$2 \int_{\Bbb R^2} X_A(\phi(x, y)) d\lambda^2(x, y) = 2 \int_{\Bbb R^2} X_{\phi^-1(A)} d\lambda^2(x, y).$$
For general measures $\theta$ on a set $\Omega$, we know the identity
$$\theta(A) = \int_{\Omega} X_A d\theta.$$
Hence, the above integral is identical to
$$2 \lambda^2(\phi^{-1}(A))$$
with $c := 2$ being our desired constant,
so
$$\sigma(A) = c \lambda^2(\phi^{-1}(A)).$$
This is identical with the definiton of the Pushforward measure, which makes $\sigma(A)$ a meausure too.
Is that a decent solution?
There are two problems with your solution: a minor one and a big one.
The minor one is that the first fundamental form (i.e. the Riemannian metric) of $E$ is, in your chosen coordinates, $\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$, therefore its determinant is $3$, not $4$.
The big one is that the problem requires you to show that equality for any global chart $\phi$, whereas you only show it for the specific chart that you construct.
There is a third problem, too, that you might not be responsible of, though: if $\sigma$ is said to be the surface-measure of $E$, why are you required to conclude that it is a measure, anymore? Alternatively, what is a "surface-measure" if not a measure? (I do not know the construction and definition of $\sigma$ that you are working with, there are several of them possible - all of them equivalent in the end.)