I have a parameter $\theta = (\theta_1, \theta_2)^\top\in\mathbb{R}^2$ with a multivariate normal prior $p(\theta) = \mathcal{N}(\theta\mid 0, I)$. The likelihood is a univariate normal distribution $\mathcal{N}(y \mid \theta_1 + \theta_2, \sigma^2)$. Is the posterior a normal distribution as well?
I know in general the product of normals is not normal. But I was wondering if in this case it is. Not sure how to check it.
Bayes' rule tells us
$$f(\theta|y)\propto f(y|\theta)f(\theta).$$
You have a single data point $y$ so that your likelihood is univariate (as clarified in the comments). Further assuming $\sigma^2$ is fixed, and letting $M\equiv (1,1)$ , your setup has likelihood and prior given by
$$y|\theta\sim N(M\theta,\sigma^2)\implies f(y|\theta)\propto \exp\left\{-\frac{1}{2\sigma^2}(y-M\theta)'(y-M\theta)\right\}\\ \theta\sim N(0,I)\implies f(\theta)\propto \exp\left\{-\frac{1}{2}\theta'\theta\right\}. $$
Now try some matrix completion of the square to write the posterior as
$$f(\theta|y)\propto \exp\left\{-\frac{1}{2}(\theta-\mu)'\Sigma^{-1}(\theta-\mu)\right\}$$
for some $\mu,\Sigma^{-1}$ (dependent on $y$ of course) to show $\theta|y\sim N(\mu,\Sigma).$